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a ) 67 , b ) 68 , c ) 87 , d ) 26 , e ) 75 | b | divide(add(add(add(add(72, 45), 72), 77), 75), divide(const_10, const_2)) | dacid obtained 72 , 45 , 72 , 77 and 75 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 72 + 45 + 72 + 77 + 75 ) / 5 = 341 / 5 = 68 . answer : b" | a = 72 + 45
b = a + 72
c = b + 77
d = c + 75
e = 10 / 2
f = d / e
|
a ) 379 , b ) 384 , c ) 345 , d ) 342 , e ) 376 | b | multiply(24, 16) | find the area of a parallelogram with base 24 cm and height 16 cm . | "area of a parallelogram = base * height = 24 * 16 = 384 cm 2 area of a parallelogram = base * height = 24 * 16 = 384 cm 2 answer : option b" | a = 24 * 16
|
a ) 18 hours , b ) 15 hours , c ) 10 hours , d ) 12 hours , e ) 8 hours | a | divide(const_1, add(divide(const_1, 30), divide(const_1, 45))) | two pipes a and b can fill a tank in 30 hours and 45 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ? | "part filled by a in 1 hour = 1 / 30 part filled by b in 1 hour = 1 / 45 part filled by ( a + b ) in 1 hour = 1 / 30 + 1 / 45 = 5 / 90 = 1 / 18 both the pipes together fill the tank in 18 hours answer is a" | a = 1 / 30
b = 1 / 45
c = a + b
d = 1 / c
|
a ) $ 12.20 , b ) $ 12.50 , c ) $ 12.55 , d ) $ 12.70 , e ) $ 13.00 | a | add(divide(add(10, 14), const_2), add(const_0_25, const_0_25)) | a vendor buys 10 t - shirts at an average price of $ 14 per t - shirt . he then buys 15 more t - shirts at an average price of $ 11 per t - shirt . what is the average price d per t - shirt that the vendor paid for these purchases ? | "correct answer : a explanation : the relevant formula for this problem is average d = ( sum ) / ( number of terms ) . another way to look at the formula is sum = average x number of terms . for the first purchase , the vendor ' s sum ( total cost ) was $ 140 , since 14 x 10 = 140 . for the second purchase , the vendor ' s cost was $ 165 , since 11 x 15 = 165 . the grand sum is then $ 140 + $ 165 , which equals $ 305 . the total number of shirts purchased was 25 , so to get the average price per shirt , we divide 305 by 25 , which equals $ 12.20 . as a result , the correct answer is a . note : a relative understanding of weighted average offers a shortcut to this problem . because the true average of 11 and 14 is 12.5 , but the vendor sells more shirts at the lower price than at the higher price , the weighted average must be less than $ 12.50 ; only answer choice a is a possibility ." | a = 10 + 14
b = a / 2
c = const_0_25 + const_0_25
d = b + c
|
a ) 2,858 , b ) 18,667 , c ) 21,429 , d ) 35,000 , e ) 65,000 | e | divide(multiply(divide(divide(multiply(add(add(const_3, 8), 12), power(const_100, const_2)), subtract(12, 8)), multiply(const_4, const_2)), const_3), const_1000) | a certain manufacturer produces items for which the production costs consist of annual fixed costs totaling $ 120,000 and variables costs averaging $ 8 per item . if the manufacturer β s selling price per item is $ 12 , how many items the manufacturer produce and sell to earn an annual profit of $ 140,000 ? | "let the items manufactured or sold bex 120000 + 8 x = 12 x - 140000 4 x = 260000 x = 65000 ans : e" | a = 3 + 8
b = a + 12
c = 100 ** 2
d = b * c
e = 12 - 8
f = d / e
g = 4 * 2
h = f / g
i = h * 3
j = i / 1000
|
a ) 25.25 s , b ) 17.45 s , c ) 35.75 s , d ) 41.45 s , e ) none of these | d | multiply(divide(24, 12), 20) | there are 20 poles with a constant distance between each pole . a car takes 24 seconds to reach the 12 th pole . how much time will it take to reach the last pole ? | let the distance between each pole be x m . then , the distance up to 12 th pole = 11 xm speed = 11 x β 24 m / s time taken to covers the total distance of 19 x = 19 x Γ 24 / 11 x = 41.45 s answer d | a = 24 / 12
b = a * 20
|
a ) 9 , b ) 19 , c ) 25 , d ) 27 , e ) 31 | c | subtract(36, subtract(add(36, 32), 57)) | in a group of 90 students , 36 are taking history , and 32 are taking statistics . if 57 students are taking history or statistics or both , then how many students are taking history but not statistics ? | number of students taking history = h = 36 number of students taking statistics = s = 32 total number of students = t = 90 number of students taking history or statistics or both = b = 57 number of students taking neither history nor statistics = n = 90 - 57 = 33 letxbe the number of students taking both history and statistics . then t = h + s + n - x or 90 = 36 + 32 + 33 - x or x = 11 now , number of students taking only history will be h - x or 36 - 11 = 25 answer : - c | a = 36 + 32
b = a - 57
c = 36 - b
|
a ) 9 , b ) 1 , c ) 2 , d ) 18 , e ) 19 | b | divide(subtract(18, power(4, 2)), 2) | if a - b = 4 and a 2 + b 2 = 18 , find the value of ab . | "2 ab = ( a 2 + b 2 ) - ( a - b ) 2 = 18 - 16 = 2 ab = 1 . answer : b" | a = 4 ** 2
b = 18 - a
c = b / 2
|
a ) 6 , b ) 8 , c ) 9 , d ) 11 , e ) 13 | b | subtract(divide(1300, 40), divide(100, 5)) | one computer can upload 100 megabytes worth of data in 5 seconds . two computers , including this one , working together , can upload 1300 megabytes worth of data in 40 seconds . how long would it take for the second computer , working on its own , to upload 100 megabytes of data ? | "since the first computer can upload 100 megabytes worth of data in 5 seconds then in 5 * 8 = 40 seconds it can upload 8 * 100 = 800 megabytes worth of data , hence the second computer in 40 seconds uploads 1300 - 800 = 500 megabytes worth of data . the second computer can upload 100 megabytes of data in 8 seconds . answer : b ." | a = 1300 / 40
b = 100 / 5
c = a - b
|
a ) 19 , b ) 29 , c ) 40 , d ) 49 , e ) 59 | c | add(subtract(88, multiply(17, 3)), 3) | a batsman makes a score of 88 runs in the 17 th inning and thus increases his averages by 3 . find his average after 17 th inning ? | "let the average after 17 th inning = x then average after 16 th inning = ( x - 3 ) therefore 16 ( x - 3 ) + 88 = 17 x therefore x = 40 answer : c" | a = 17 * 3
b = 88 - a
c = b + 3
|
a ) 2 / 5 , b ) 1 / 6 , c ) 1 / 4 , d ) 1 / 9 , e ) 2 / 6 | d | multiply(2, add(divide(const_1, 30), divide(const_1, 45))) | two persons a and b can complete a piece of work in 30 days and 45 days respectively . if they work together , what part of the work will be completed in 2 days ? | "a ' s one day ' s work = 1 / 30 b ' s one day ' s work = 1 / 45 ( a + b ) ' s one day ' s work = 1 / 30 + 1 / 45 = 1 / 18 the part of the work completed in 2 days = 2 ( 1 / 18 ) = 1 / 9 . answer d" | a = 1 / 30
b = 1 / 45
c = a + b
d = 2 * c
|
a ) 11 pi , b ) 12 pi , c ) 13 pi , d ) 14 pi , e ) 15 pi | b | multiply(add(divide(add(11, const_3), const_100), const_3), subtract(add(add(power(divide(4, 2), const_2), divide(2, 2)), 11), subtract(add(power(divide(4, 2), const_2), divide(2, 2)), 1))) | find the area between two concentric circles defined by x 2 + y 2 - 2 x + 4 y + 1 = 0 x 2 + y 2 - 2 x + 4 y - 11 = 0 | rewrite equations of circles in standard form . hence equation x 2 + y 2 - 2 x + 4 y + 1 = 0 may be written as ( x - 1 ) 2 + ( y + 2 ) 2 = 4 = 22 and equation x 2 + y 2 - 2 x + 4 y - 11 = 0 as ( x - 1 ) 2 + ( y + 2 ) 2 = 16 = 42 knowing the radii , the area of the ring is pi ( 4 ) 2 - pi ( 2 ) 2 = 12 pi correct answer is b ) 12 pi | a = 11 + 3
b = a / 100
c = b + 3
d = 4 / 2
e = d ** 2
f = 2 / 2
g = e + f
h = g + 11
i = 4 / 2
j = i ** 2
k = 2 / 2
l = j + k
m = l - 1
n = h - m
o = c * n
|
a ) 9 , b ) 8 , c ) 10 , d ) 8.5 , e ) 21 | e | divide(63, 3) | stacy has a 63 page history paper due in 3 days . how many pages per day would she have to write to finish on time ? | "63 / 3 = 21 answer : e" | a = 63 / 3
|
a ) 90 % , b ) 30 % , c ) 40 % , d ) 70 % , e ) 80 % | a | subtract(100, 10) | john want to buy a $ 100 trouser at the store , but he think it β s too expensive . finally , it goes on sale for $ 10 . what is the percent decrease ? | "the is always the difference between our starting and ending points . in this case , it β s 100 β 10 = 90 . the β original β is our starting point ; in this case , it β s 100 . ( 90 / 100 ) * 100 = ( 0.9 ) * 100 = 90 % . a" | a = 100 - 10
|
a ) 3000 m 3 , b ) 4580 m 3 , c ) 18500 m 3 , d ) 4900 m 3 , e ) 4700 m 3 | a | divide(multiply(multiply(2, 45), multiply(2, const_1000)), multiply(const_1, const_60)) | a river 2 m deep and 45 m wide is flowing at the rate of 2 kmph the amount of water that runs into the sea per minute is ? | "explanation : ( 2000 * 2 * 45 ) / 60 = 3000 m 3 answer : option a" | a = 2 * 45
b = 2 * 1000
c = a * b
d = 1 * const_60
e = c / d
|
a ) 338 , b ) 277 , c ) 342 , d ) 882 , e ) 304 | e | subtract(subtract(400, divide(multiply(400, 20), const_100)), divide(multiply(subtract(400, divide(multiply(400, 20), const_100)), 5), const_100)) | the sale price sarees listed for rs . 400 after successive discount is 20 % and 5 % is ? | "400 * ( 80 / 100 ) * ( 95 / 100 ) = 304 answer : e" | a = 400 * 20
b = a / 100
c = 400 - b
d = 400 * 20
e = d / 100
f = 400 - e
g = f * 5
h = g / 100
i = c - h
|
a ) 1310 , b ) 1320 , c ) 1322 , d ) 1323 , e ) 1312 | b | multiply(const_3, multiply(11, multiply(5, 8))) | calculate how many seconds it eill take for 4 bells to toll together again , given that they begin to toll together respectively at the intervals of 5 , 8 , 11 and 15 seconds . ? | lcm of 5 , 8 , 11 and 15 is 1320 lcm = 1320 answer : b | a = 5 * 8
b = 11 * a
c = 3 * b
|
a ) 1300 , b ) 1100 , c ) 1000 , d ) 1200 , e ) 800 | d | multiply(divide(2250, add(multiply(divide(subtract(12, 6), subtract(3, 1)), 3), 6)), 8) | the price of 3 pants and 6 t - shirts is rs . 2250 . with the same money one can buy 1 pant and 12 t - shirts . if one wants to buy 8 t - shirts , how much shall she have to pay ? | "let the price of a pant and a t - shirt be rs . x and rs . y respectively . then , 3 x + 6 y = 2250 . . . . ( i ) and x + 12 y = 2250 . . . . ( ii ) divide equation ( i ) by 3 , we get the below equation . = x + 2 y = 750 . - - - ( iii ) now subtract ( iii ) from ( ii ) x + 12 y = 2250 ( - ) x + 2 y = 750 - - - - - - - - - - - - - - - - 10 y = 1500 - - - - - - - - - - - - - - - - y = 150 cost of 8 t - shirts = 8 * 150 = 1200 answer : d" | a = 12 - 6
b = 3 - 1
c = a / b
d = c * 3
e = d + 6
f = 2250 / e
g = f * 8
|
a ) 200 m , b ) 250 m , c ) 300 m , d ) 350 m , e ) 400 m | a | multiply(divide(multiply(30, const_1000), const_3600), 24) | a train running at the speed of 30 km / hr crosses a pole in 24 seconds . what is the length of the train ? | "speed = ( 30 * 5 / 18 ) m / sec = ( 25 / 3 ) m / sec length of the train = ( speed x time ) = ( 25 / 3 * 24 ) m = 200 m . answer : a" | a = 30 * 1000
b = a / 3600
c = b * 24
|
a ) 100 , b ) 210 , c ) 300 , d ) 400 , e ) 500 | b | subtract(800, add(add(190, subtract(500, 190)), subtract(400, subtract(500, 190)))) | - - - - - - - - - - - - - - - - yes - - - - - - - - - no - - - - unsure subject m - - - - 500 - - - - - - - - 200 - - - - - 100 subject r - - - - 400 - - - - - - - - 100 - - - - - 300 a total of 800 students were asked whether they found two subjects , m and r , interesting . each answer was either yes or no or unsure , and the numbers of students who gave these answers are listed in the table above . if 190 students answered yes only for subject m , how many of the students did not answer yes for either subject ? | "since 190 students answered yes only for subject m , then the remaining 310 students who answered yes for subject m , also answered yes for subject r . so , 310 students answered yes for both subjects . if 310 students answered yes for both subjects , then 400 - 310 = 90 students answered yes only for subject r . so , we have that : 200 students answered yes only for subject m ; 90 students answered yes only for subject r ; 300 students answered yes for both subjects ; therefore 800 - ( 200 + 90 + 300 ) = 210 students did not answer yes for either subject . answer : b ." | a = 500 - 190
b = 190 + a
c = 500 - 190
d = 400 - c
e = b + d
f = 800 - e
|
a ) 50 % , b ) 68 % , c ) 72 % , d ) 76 % , e ) 80 % | a | multiply(divide(add(const_100, 20), add(add(const_100, 20), const_100)), const_100) | in may mrs lee ' s earnings were 45 percent of the lee family ' s total income . in june mrs lee earned 20 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs lee ' s earnings were approximately what percent of the lee family ' s total income ? | "let in may lee family ' s total income = 100 in may mrs lee ' s income = 45 in may rest of the family ' s income = 55 in june mrs lees income = 45 * 120 / 100 = 54 in june total income = 54 + 55 = 109 % of mrs lee ' s income = 54 / 109 = 49.54 ( a )" | a = 100 + 20
b = 100 + 20
c = b + 100
d = a / c
e = d * 100
|
a ) 70578 , b ) 82578 , c ) 80578 , d ) 80568 , e ) 89045 | c | multiply(217, 217) | 217 * 217 + 183 * 183 | "explanation : 1 / 2 Γ 2 ( a 2 + b 2 ) = 1 / 2 Γ [ ( a + b ) 2 + ( a β b ) 2 ] = 1 / 2 Γ [ ( 217 + 183 ) 2 + ( 217 β 183 ) 2 ] = 80578 option c" | a = 217 * 217
|
a ) 225 , b ) 245 , c ) 275 , d ) 325 , e ) 375 | c | subtract(multiply(divide(multiply(45, const_1000), const_3600), 30), 100) | the length of a bridge in meters , which a train 100 - meters long and traveling at 45 km / hr can cross in 30 seconds is ? | "45 km / h = 45000 m / 3600 s = 12.5 m / s in 30 seconds , the train can go 30 ( 12.5 ) = 375 meters let x be the length of the bridge . x + 100 = 375 meters x = 275 meters the answer is c ." | a = 45 * 1000
b = a / 3600
c = b * 30
d = c - 100
|
a ) 8.62 % , b ) 8.88 % , c ) 8.32 % , d ) 8 % , e ) 9 % | b | sqrt(divide(632, divide(800, const_100))) | reema took a loan of rs 800 with simple interest for as many years as the rate of interest . if she paid rs . 632 as interest at the end of the loan period , what was the rate of interest . | "explanation : let rate = r % then time = r years . = > 800 β r β r / 100 = 632 = > r 2 = 79 = > r = 8.88 % option b" | a = 800 / 100
b = 632 / a
c = math.sqrt(b)
|
a ) 4 minutes , b ) 32 / 7 minutes , c ) 192 / 7 minutes , d ) 224 / 7 minutes , e ) 32 / 5 minutes | e | inverse(add(divide(const_1, 32), add(divide(const_1, 32), divide(3, 32)))) | pipe a can fill a tank in 32 minutes . pipe b can fill the same tank 3 times faster than pipe a . if both the pipes are connected to the tank so that they fill the tank simultaneously , how long will it take for the empty tank to overflow ? | pipe a can fill a tank in 32 minutes - - > the rate of a = 1 / 32 tank / minute ; pipe b can fill the same tank 3 times fasterthan pipe a - - > the rate of b = 1 / 32 + 3 / 32 = 4 / 32 tank / minute . their combined rate = 1 / 32 + 4 / 32 = 5 / 32 tank / minute ; therefore , a and b can fill the tank in 32 / 5 minutes . answer : e | a = 1 / 32
b = 1 / 32
c = 3 / 32
d = b + c
e = a + d
f = 1/(e)
|
a ) 49 km / hr , b ) 50 km / hr , c ) 45 km / hr , d ) 65 km / hr , e ) 87 km / hr | a | divide(divide(subtract(125, multiply(multiply(4, const_0_2778), 4)), 4), const_0_2778) | a train 125 m long passes a man , running at 4 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is : | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 4 ) km / hr . x - 4 = 45 = = > x = 49 km / hr . answer : a" | a = 4 * const_0_2778
b = a * 4
c = 125 - b
d = c / 4
e = d / const_0_2778
|
a ) $ 500 , b ) $ 800 , c ) $ 1000 , d ) $ 200 , e ) $ 750 | b | divide(1000, 1) | divide $ 1000 among a , b in the ratio 1 : 4 . how many $ that b get ? | "sum of ratio terms = 1 + 4 = 5 a = 1000 * 4 / 5 = $ 800 answer is b" | a = 1000 / 1
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a ) 2 , b ) 4 , c ) 8 , d ) 3 , e ) 1 | c | subtract(multiply(multiply(divide(3, add(5, 3)), add(const_10, 16)), 3), multiply(divide(5, add(5, 3)), add(const_10, 16))) | an alloy of zinc and copper contains the metals in the ratio 5 : 3 . the quantity of zinc to be added to 16 kg of the alloy so that the ratio of the metal may be 3 : 1 is : | "5 / 8 * 16 = 10 3 / 8 * 16 = 6 ( 10 + x ) / 6 = 3 / 1 = 8 answer : c" | a = 5 + 3
b = 3 / a
c = 10 + 16
d = b * c
e = d * 3
f = 5 + 3
g = 5 / f
h = 10 + 16
i = g * h
j = e - i
|
a ) 1 / 62 , b ) 1 / 64 , c ) 1 / 65 , d ) 1 / 66 , e ) 1 / 67 | b | divide(multiply(add(add(const_100, const_60), const_1), 2), const_100) | what is the value of ( 2 ) ^ - 6 ? | "2 ^ - 6 = 1 / ( 2 ) ^ 6 = 1 / 64 answer : b" | a = 100 + const_60
b = a + 1
c = b * 2
d = c / 100
|
a ) 10000 , b ) 12000 , c ) 13000 , d ) 10860 , e ) 14400 | d | multiply(7800, power(add(const_1, divide(18, const_100)), 2)) | the population of a village is 7800 . it increases annually at the rate of 18 % p . a . what will be its population after 2 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 7800 Γ 118 / 100 Γ 118 / 100 = 10860 d" | a = 18 / 100
b = 1 + a
c = b ** 2
d = 7800 * c
|
a ) 4 , 3,22 , b ) 4 , 4,22 , c ) 10 , 4,8 , d ) 9 , 6,12 , e ) 9 , 2,23 | c | divide(multiply(4, 5), 5) | find the numbers which are in the ratio 5 : 2 : 4 such that the sum of the first and the second added to the difference of the third and the second is 18 ? | "let the numbers be a , b and c . a : b : c = 5 : 2 : 4 given , ( a + b ) + ( c - b ) = 21 = > a + c = 21 = > 5 x + 4 x = 18 = > x = 2 a , b , c are 5 x , 2 x , 4 x a , b , c are 10 , 4 , 8 . answer : c" | a = 4 * 5
b = a / 5
|
a ) 3.69 , b ) 3.66 , c ) 3.6 , d ) 100 , e ) 3.61 | d | multiply(divide(subtract(168, multiply(subtract(33, 30), divide(subtract(186, 168), subtract(36, 33)))), 30), 20) | apple costs l rupees per kilogram for first 30 kgs and q rupees per kilogram for each additional kilogram . if the price of 33 kilograms is 168 and for 36 kgs of apples is 186 then the cost of first 20 kgs of apples is | "ans : by framing equations we get 30 l + 3 q = 168 30 l + 6 q = 186 eliminate q by multiplying the first equation by 2 and subtracting second equation from the first 30 l = 150 = > l = 5 then we get l = 5 cost of 20 kgs of apples = 20 x 5 = 100 answer : d" | a = 33 - 30
b = 186 - 168
c = 36 - 33
d = b / c
e = a * d
f = 168 - e
g = f / 30
h = g * 20
|
a ) 5.6 sec , b ) 8.9 sec , c ) 10.8 sec , d ) 12 sec , e ) 15 sec | d | divide(add(200, 160), multiply(add(68, 40), const_0_2778)) | two trains 200 m and 160 m long run at the speed of 68 kmph and 40 kmph in opposite directions in parallel tracks . the time which they take to cross each other is ? | "relative speed = 68 + 40 = 108 kmph * 5 / 18 = 30 m / s distance covered in crossing each other = 200 + 160 = 360 m required time = 360 * 1 / 30 = 12 sec answer is d" | a = 200 + 160
b = 68 + 40
c = b * const_0_2778
d = a / c
|
a ) 16 , b ) 12 , c ) 11 , d ) 20 , e ) 10 | c | multiply(11, const_1) | the total age of a and b is 11 years more than the total age of b and c . c is how many years younger than a . ? | "( a + b ) - ( b - c ) = 11 a - c = 11 answer is c" | a = 11 * 1
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a ) 0.23255 , b ) 0.14544 , c ) 0.25632 , d ) 0.35466 , e ) 0.63435 | a | multiply(divide(divide(480, divide(60, const_100)), add(multiply(multiply(3, const_100), const_1000), multiply(add(multiply(const_4, const_10), const_4), const_1000))), const_100) | lagaan is levied on the 60 percent of the cultivated land . the revenue department collected total rs . 3 , 44,000 through the lagaan from the village of mutter . mutter , a very rich farmer , paid only rs . 480 as lagaan . the percentage of total land of mutter over the total taxable land of the village is : | total land of sukhiya = \ inline \ frac { 480 x } { 0.6 } = 800 x \ therefore cultivated land of village = 344000 x \ therefore required percentage = \ inline \ frac { 800 x } { 344000 } \ times 100 = 0.23255 a | a = 60 / 100
b = 480 / a
c = 3 * 100
d = c * 1000
e = 4 * 10
f = e + 4
g = f * 1000
h = d + g
i = b / h
j = i * 100
|
a ) 0.15 , b ) 0.20 , c ) 0.25 , d ) 0.30 , e ) 0.33 | b | divide(const_4, multiply(const_4, 5)) | s = { 2 , 3 , 4 , 5 } b = { 4 , 5 , 6 , 7 , 8 } two integers will be randomly selected from the sets above , one integer from set s and one integer from set b . what is the probability that the sum of the two integers will equal 9 ? | the total number of pairs s , b possible is 4 * 5 = 20 . out of these 20 pairs only 4 sum up to 9 : ( 2 , 7 ) ; ( 3 , 6 ) , ( 4 , 5 ) and ( 5 , 4 ) . the probability thus is 4 / 20 = 0.2 . answer : b . | a = 4 * 5
b = 4 / a
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a ) 3 / 20 , b ) 1 / 3 , c ) 5 / 2 , d ) 2 / 7 , e ) 4 / 5 | e | divide(multiply(3, 3), add(3, multiply(3, 3))) | every student in a room is either a junior or a senior . there is at least one junior and at least one senior in the room . if 1 / 3 of the juniors is equal to 4 / 3 of the seniors , what fraction of the students in the room are juniors ? | "let total number of juniors = j total number of seniors = s ( 1 / 3 ) j = ( 4 / 3 ) s = > s = 1 / 4 j total number of students = j + s = ( 5 / 4 ) j fraction of the students in the room are juniors = j / ( j + s ) = j / [ ( 5 / 4 ) j ] = 4 / 5 answer e" | a = 3 * 3
b = 3 * 3
c = 3 + b
d = a / c
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a ) 18 , b ) 20 , c ) 22 , d ) 23 , e ) 24 | d | divide(130, multiply(7, 1)) | how many 7 in between 1 to 130 ? | "7 , 17,27 , 37,47 , 57,67 , 70,71 , 72,73 , 74,75 , 76,77 ( two 7 ' s ) , 78 , 79,87 , 97,107 , 117,127 23 7 ' s between 1 to 130 answer : d" | a = 7 * 1
b = 130 / a
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a ) 1.2 , b ) 1.45 , c ) 2.2 , d ) 2.9 , e ) none | d | divide(multiply(subtract(100.7, const_100), const_2), subtract(100.48, const_100)) | given that 100.48 = x , 100.70 = y and xz = y Β² , then the value of z is close to | sol . xz = y Β² β ( 10 0.48 ) z ( 100.70 ) 2 ( 10 0.48 z ) = 10 ( 2 Γ 0.70 ) = 101.40 β 0.48 z = 1.40 β z = 140 / 48 = 35 / 12 = 2.9 ( approx ) . answer d | a = 100 - 7
b = a * 2
c = 100 - 48
d = b / c
|
a ) 4 , b ) 8 , c ) 12 , d ) 16 , e ) 20 | b | divide(add(multiply(8, 6), multiply(8, 5)), add(6, 5)) | the number of stamps that p and q had were in the ratio of 7 : 4 respectively . after p gave q 8 stamps , the ratio of the number of p ' s stamps to the number of q ' s stamps was 6 : 5 . as a result of the gift , p had how many more stamps than q ? | p started with 7 k stamps and q started with 4 k stamps . ( 7 k - 8 ) / ( 4 k + 8 ) = 6 / 5 35 k - 24 k = 48 + 40 11 k = 88 k = 8 p has 7 ( 8 ) - 8 = 48 stamps and q has 4 ( 8 ) + 8 = 40 stamps . the answer is b . | a = 8 * 6
b = 8 * 5
c = a + b
d = 6 + 5
e = c / d
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a ) 50 , b ) 80 , c ) 100 , d ) 125 , e ) 21 | e | multiply(divide(subtract(power(add(divide(multiply(10, const_2), const_1000), const_3), const_2), const_4), const_4), const_100) | a circular logo is enlarged to fit the lid of a jar . the new diameter is 10 per cent larger than the original . by what percentage has the area of the logo increased ? | "let old diameter be 4 , so radius is 2 old area = 4 Ο new diameter is 4.4 , so radius is 2.2 new area = 4.84 Ο increase in area is 0.84 Ο % increase in area = 0.84 / 4 * 100 so , % increase is 21 % answer will be ( e )" | a = 10 * 2
b = a / 1000
c = b + 3
d = c ** 2
e = d - 4
f = e / 4
g = f * 100
|
a ) 20 , b ) 18 , c ) 21.4 , d ) 16 , e ) 10 | c | divide(add(multiply(25, const_10), divide(multiply(25, const_10), const_2)), add(divide(multiply(25, const_10), 20), divide(divide(multiply(25, const_10), const_2), 25))) | a certain car traveled twice as many miles from town a to town b as it did from town b to town c . from town a to town b , the car averaged 20 miles per gallon , and from town b to town c , the car averaged 25 miles per gallon . what is the average miles per gallon that the car achieved on its trip from town a through town b to town c ? | "step 1 ) took lcm of 20 and 25 . . came as 100 . step 2 ) 100 distance between b to c . . . do 100 / 25 hence 4 gallons used step 3 ) twice distance . . hence 100 * 2 = 200 . . . do as above . . 200 / 20 = 10 gallons used step 4 ) total gallons . . 4 + 10 = 14 gallons step ) total miles = 100 + 200 = 300 miles hence . . average of whole journey = 300 / 14 which comes to 21.4 answer : c" | a = 25 * 10
b = 25 * 10
c = b / 2
d = a + c
e = 25 * 10
f = e / 20
g = 25 * 10
h = g / 2
i = h / 25
j = f + i
k = d / j
|
a ) $ 8,000 , b ) $ 5,600 , c ) $ 3,200 , d ) $ 2,400 , e ) $ 4,800 | e | subtract(multiply(multiply(const_100, const_10), multiply(const_2, const_4)), multiply(multiply(multiply(const_100, 20), multiply(const_2, const_4)), multiply(divide(20, const_100), const_2))) | the market value of a certain machine decreased by 20 percent of its purchase price each year . if the machine was purchased in 1982 for its market value of $ 8,000 , what was its market value two years later ? | "e . market value in 1982 = $ 8000 market value in 1983 = $ 8000 - ( $ 8000 x 20 / 100 ) = 8000 - 1600 = $ 6400 market value in 1984 = market value in 1983 - ( 20 % of $ 8000 ) = 6400 - 1600 = $ 4800" | a = 100 * 10
b = 2 * 4
c = a * b
d = 100 * 20
e = 2 * 4
f = d * e
g = 20 / 100
h = g * 2
i = f * h
j = c - i
|
a ) 20 , b ) 24 , c ) 26 , d ) 23 , e ) 30 | d | sqrt(add(multiply(131, const_2), 267)) | sum of the squares of 3 no . ' s is 267 and the sum of their products taken two at a time is 131 . find the sum ? | "( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 267 + 2 * 131 a + b + c = β 529 = 23 d" | a = 131 * 2
b = a + 267
c = math.sqrt(b)
|
a ) 30 , b ) 26 , c ) 28 , d ) 29 , e ) 45 | b | subtract(add(add(multiply(10, 15), 16), 10), multiply(10, 15)) | the average age of 10 students in a class is 15 years . if the age of teacher is also included , the average becomes 16 years , find the age of the teacher . | "explanation : if teacher ' s age is 15 years , there is no change in the average . but teacher has contributed 1 year to all the students along with maintaining his age at 16 . age of teacher = average age of all + total increase in age = 16 + ( 1 x 10 ) = 26 years answer : b" | a = 10 * 15
b = a + 16
c = b + 10
d = 10 * 15
e = c - d
|
a ) 20 % , b ) 25 % , c ) 40 % , d ) 60 % , e ) 75 % | b | multiply(subtract(const_1, divide(divide(60, multiply(multiply(const_2, const_5), multiply(const_2, const_5))), subtract(const_1, divide(20, multiply(multiply(const_2, const_5), multiply(const_2, const_5)))))), multiply(multiply(const_2, const_5), multiply(const_2, const_5))) | the maitre ' d at an expensive manhattan restaurant has noticed that 60 % of the couples order dessert and coffee . however , 20 % of the couples who order dessert do n ' t order coffee . what is the probability w that the next couple the maitre ' d seats will not order dessert ? | "could you use a venn diagram and just go with the number 100 . 60 people order dessert and coffee . . . which is the union of d and c . w = 2 / 10 of d are n ' t in d u c = so 8 / 10 of d are in duc which means = 60 = 8 / 10 d . so d in total = 75 , and 15 d ' s are n ' t in d union c . which means 25 people are in c only + neither . b 25 %" | a = 2 * 5
b = 2 * 5
c = a * b
d = 60 / c
e = 2 * 5
f = 2 * 5
g = e * f
h = 20 / g
i = 1 - h
j = d / i
k = 1 - j
l = 2 * 5
m = 2 * 5
n = l * m
o = k * n
|
a ) 11 , b ) 444 , c ) 2881 , d ) 287 , e ) 456 | e | add(multiply(divide(60, subtract(21, 17)), 17), multiply(divide(60, subtract(21, 17)), 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 17 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 = 17 + 21 = 38 t = 12 d = 38 * 12 = 456 answer : e" | a = 21 - 17
b = 60 / a
c = b * 17
d = 21 - 17
e = 60 / d
f = e * 21
g = c + f
|
a ) 56 , b ) 86.7 , c ) 25.7 , d ) 66 , e ) 64.7 | e | add(add(multiply(100, divide(70, 200)), multiply(divide(50, 200), 200)), multiply(divide(85, 200), 120)) | in a college , 70 % of total 100 arts students are locals . 50 % of students from 200 science students are locals and 85 % of total 120 commerce students are locals . what is the total percentage of locals from arts , science and commerce . | "locals from arts = 70 % of 100 = 70 locals from science = 50 % of 200 = 100 locals from commerce = 85 % of 120 = 102 total local percentage = ( 70 + 100 + 102 ) / ( 100 + 200 + 120 ) = 272 / 420 = 0.647 e" | a = 70 / 200
b = 100 * a
c = 50 / 200
d = c * 200
e = b + d
f = 85 / 200
g = f * 120
h = e + g
|
a ) 120 , b ) 123 , c ) 126 , d ) 129 , e ) 132 | b | multiply(add(add(subtract(multiply(2, const_10), 2), 1), multiply(2, const_10)), 3) | every letter in the alphabet has a number value that is equal to its place in the alphabet . thus , the letter a has a value of 1 , the letter b has a value of 2 , the letter c has a value of 3 , etc . . . the number value of a word is obtained by adding up the value of the letters in the word and then multiplying that sum by the length of the word . what is the number value of the word ` ` cow ' ' ? | "` ` cow ' ' = ( 3 + 15 + 23 ) * 3 = 123 . the answer is b ." | a = 2 * 10
b = a - 2
c = b + 1
d = 2 * 10
e = c + d
f = e * 3
|
a ) 120 , b ) 80 , c ) 100 , d ) 150 , e ) 140 | a | multiply(divide(const_60, 1), 60) | if the population of a certain country increases at the rate of 1 person every 60 seconds , by how many persons does the population increase in 2 hours ? | "answer = 1 * 120 ( 2 hour = 120 minutes ) = 120 answer = a" | a = const_60 / 1
b = a * 60
|
a ) 11 , b ) - 10.83 , c ) 71 , d ) 20.83 , e ) - 12 | b | divide(subtract(14, 79), 6) | a straight line in the xy - plane has y - intercept of 79 and slope of 6 . on this line the y - coordinate of the point is 14 , then what is the x - coordinate of the point ? | "eq of line = y = mx + c c = 79 m = 6 y = 14 substitute given : x = ( y - c ) / m = ( 14 - 79 ) / 6 = - 65 / 6 = - 10.83 correct option is b" | a = 14 - 79
b = a / 6
|
a ) 400 , b ) 1550 , c ) 1575 , d ) 1600 , e ) 1625 | c | add(rectangle_area(200, 5), rectangle_area(120, 5)) | a rectangular lawn of length 200 m by 120 m has two roads running along its center , one along the length and the other along the width . if the width of the roads is 5 m what is the area t covered by the two roads ? | "area covered by road along the length = 5 * 200 = 1000 square meter area covered by road along the width = 5 * 120 = 600 square meter common area in both roads ( where the roads intersect ) = square with side 5 meter = 5 * 5 = 25 total area of the roads t = 1000 + 600 - 25 = 1575 answer : option c" | a = rectangle_area + (
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a ) 45 % , b ) 48 % , c ) 57 % , d ) 59 % , e ) 61 % | b | multiply(divide(11628, add(add(5136, 7636), 11628)), const_100) | 3 candidates in an election and received 5136 , 7636 and 11628 votes respectively . what % of the total votes did the winningcandidate got in that election ? | "total number of votes polled = ( 5136 + 7636 + 11628 ) = 24400 so , required percentage = 11628 / 24400 * 100 = 48 % b" | a = 5136 + 7636
b = a + 11628
c = 11628 / b
d = c * 100
|
a ) 2984 , b ) 2983 , c ) 3500 , d ) 2981 , e ) none of these | c | subtract(3550, divide(1002, 20.04)) | 3550 - ( 1002 / 20.04 ) = ? | "3550 - 50 = 3500 answer : c" | a = 1002 / 20
b = 3550 - a
|
a ) z / 2 , b ) z , c ) 3 z / 2 , d ) 2 z , e ) 3 z | e | multiply(const_2, add(const_1, divide(const_1, const_2))) | two ants , arthur and amy , have discovered a picnic and are bringing crumbs back to the anthill . amy makes twice as many trips and carries one and a half times as many crumbs per trip as arthur . if arthur carries a total of z crumbs to the anthill , how many crumbs will amy bring to the anthill , in terms of z ? | "lets do it by picking up numbers . let arthur carry 2 crumbs per trip , this means amy carries 3 crumbs per trip . also let arthur make 2 trips and so amy makes 4 trips . thus total crumbs carried by arthur ( z ) = 2 x 2 = 4 , total crumbs carried by amy = 3 x 4 = 12 . 12 is 3 times 4 , so e" | a = 1 / 2
b = 1 + a
c = 2 * b
|
a ) 135 , b ) 227 , c ) 268 , d ) 198 , e ) 176 | a | multiply(divide(435, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | a , b and c rents a pasture for rs . 435 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should c pay ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 9 / 29 * 435 = 135 answer : a" | a = 12 * 8
b = 16 * 9
c = a + b
d = 18 * 6
e = c + d
f = 435 / e
g = 16 * 9
h = f * g
|
a ) 3 / 7 , b ) 5 / 4 , c ) 5 / 8 , d ) 5 / 9 , e ) 6 / 9 | a | divide(divide(3, 4), add(divide(3, 4), const_1)) | a waiter ' s salary consists of his salary and tips . during one week his tips were 3 / 4 of his salary . what fraction of his income came from tips ? | "income = salary ( s ) + tips = s + s * 3 / 4 = s * 7 / 4 tips = s * 3 / 4 fraction of his income came from tips = ( s * 3 / 4 ) / ( s * 7 / 4 ) = 3 / 7 answer : a" | a = 3 / 4
b = 3 / 4
c = b + 1
d = a / c
|
a ) a ) 2 , b ) b ) 3 , c ) c ) 5 , d ) d ) 5 , e ) e ) 6 | c | subtract(509, multiply(add(multiply(add(const_4, const_1), const_10), add(const_4, const_2)), 9)) | the least number which must be subtracted from 509 to make it exactly divisible by 9 is : | on dividing 509 by 9 , we get remainder = 5 therefore , required number to be subtracted = 5 answer : c | a = 4 + 1
b = a * 10
c = 4 + 2
d = b + c
e = d * 9
f = 509 - e
|
a ) 8 , b ) 7 , c ) 10 , d ) 9 , e ) 6 | a | divide(add(9, 8), subtract(9, 8)) | if x / y = 9 / 8 , then ( 8 x + 7 y ) / ( 8 x Γ’ β¬ β 7 y ) = ? | "answer dividing numerator as well as denominator by y , we get given exp . = ( 8 x + 7 y ) / ( 8 x Γ’ β¬ β 7 y ) = ( 8 x / y + 7 ) / ( 8 x / y Γ’ β¬ β 7 ) since x / y = 9 / 8 this implies that = [ ( 8 * 9 ) / 8 + 7 ] / [ ( 8 * 9 ) / 8 - 7 ) ] = ( 9 + 7 ) / ( 9 - 7 ) = 8 option : a" | a = 9 + 8
b = 9 - 8
c = a / b
|
a ) 15 sec , b ) 16 sec , c ) 18 sec , d ) 11 sec , e ) 8 sec | e | multiply(divide(140, multiply(63, const_1000)), const_3600) | a train 140 m long , running with a speed of 63 km / hr will pass a tree in ? | "speed = 63 * 5 / 18 = 35 / 2 m / sec time taken = 140 * 2 / 35 = 8 sec answer : e" | a = 63 * 1000
b = 140 / a
c = b * 3600
|
a ) 100 rs , b ) 150 rs , c ) 160 rs , d ) 180 rs , e ) 200 rs | c | divide(subtract(divide(multiply(multiply(8000, 6), 2), const_100), divide(multiply(multiply(8000, 4), 2), const_100)), 2) | a person borrows rs . 8000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 6 p . a for 2 years . find his gain in the transaction per year . | "gain in 2 years = [ ( 8000 * 6 * 2 ) / 100 ] - [ ( 8000 * 4 * 2 ) / 100 ] 960 - 640 = 320 gain in 1 year = ( 320 / 2 ) = 160 rs answer : c" | a = 8000 * 6
b = a * 2
c = b / 100
d = 8000 * 4
e = d * 2
f = e / 100
g = c - f
h = g / 2
|
a ) 20,20 , b ) 20,10 , c ) 25,9 , d ) 30,10 , e ) none of these | c | subtract(add(divide(multiply(16, 5), subtract(5, const_1)), 5), 16) | the ages of two persons differ by 16 years . if 5 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively | "explanation : let their ages be x and ( x + 16 ) years . 5 ( x - 5 ) = ( x + 16 - 5 ) or 4 x = 36 or x = 9 . their present ages are 25 years and 9 years option c" | a = 16 * 5
b = 5 - 1
c = a / b
d = c + 5
e = d - 16
|
a ) 2.5 % , b ) 17.67 % , c ) 28.3 % , d ) 45.2 % , e ) 73.6 % | c | multiply(divide(multiply(divide(30, const_100), subtract(1, divide(1, 3))), subtract(1, divide(1, 3))), const_100) | a library branch originally contained 18360 volumes , 30 % of which were fiction novels . 1 / 3 of the volumes were transferred to another location and 1 / 3 of the volumes transferred were fiction novels . what percent of the remaining collection was fiction novels ? | "as everything is either fraction or percentage , the given figure 18360 is just a false alarm . we can do this by assuming that originally the branch had 100 volumes . originally : total - 100 fiction - 30 transferred : total - 33 ( one third of original total ) fiction - 11 ( one third of those transferred ) remaining : total - 100 β 33 = 67100 β 33 = 67 fiction - 30 β 11 = 1930 β 11 = 19 to find : 19 is what percent of 67 28.3 option c" | a = 30 / 100
b = 1 / 3
c = 1 - b
d = a * c
e = 1 / 3
f = 1 - e
g = d / f
h = g * 100
|
a ) 11 am . , b ) 6 p . m . , c ) 3 pm . , d ) 2 p . m . , e ) 1 p . m . | b | add(divide(add(2, 2), add(1, 2)), 1000) | the distance between 2 cities a and b is 1000 km . a train starts from a at 1 p . m . and travels towards b at 100 km / hr . another starts from b at 2 p . m . and travels towards a at 150 km / hr . at what time do they meet ? | "suppose they meet x hrs after 1 p . m . distance moved by first in x hrs + distance moved by second in ( x - 1 ) hrs = 1000 100 x + 150 ( x - 1 ) = 1000 x = 4.60 = 5 hrs they meet at 1 + 5 = 6 p . m . answer is b" | a = 2 + 2
b = 1 + 2
c = a / b
d = c + 1000
|
a ) 12 , b ) 10 , c ) 9 , d ) 8 , e ) 11 | d | divide(subtract(subtract(subtract(subtract(50, const_1), const_2), const_3), const_4), 5) | in a certain brick wall , each row of bricks above the bottom row contains one less brick than the row just below it . if there are 5 rows in all and a total of 50 bricks in the wall , how many bricks does the bottom row contain ? | the bottom row has x bricks x + x - 1 + x - 2 + x - 3 + x - 4 = 50 5 x - 10 = 50 5 x = 40 x = 8 answer : d | a = 50 - 1
b = a - 2
c = b - 3
d = c - 4
e = d / 5
|
a ) 8 , b ) 3 , c ) 6 , d ) 7 , e ) 9 | c | divide(multiply(4, 4), const_10) | if 17 ^ 4 * 13 ^ 4 * 9 ^ 4 * 8 ^ 4 = x , what is the units digit of x ? | "17 ^ 4 * 13 ^ 4 * 9 ^ 4 * 8 ^ 4 = x units digit of x will be 7 ^ 4 * 3 ^ 4 * 9 ^ 4 * 8 ^ 4 = 1 * 1 * 1 * 6 = 6 therefore , units digit of x = 6 answer c" | a = 4 * 4
b = a / 10
|
a ) $ 1.75 , b ) $ 1.85 , c ) $ 2.00 , d ) $ 2.15 , e ) $ 2.30 | a | add(divide(subtract(8.25, multiply(divide(multiply(subtract(6, const_1), 6), const_2), 0.15)), 6), multiply(subtract(6, const_1), 0.15)) | mark bought a set of 6 flower pots of different sizes at a total cost of $ 8.25 . each pot cost 0.15 more than the next one below it in size . what was the cost , in dollars , of the largest pot ? | "this question can be solved with a handful of different algebra approaches ( as has been shown in the various posts ) . since the question asks for the price of the largest pot , and the answers are prices , we can test the answers . we ' re told that there are 6 pots and that each pot costs 25 cents more than the next . the total price of the pots is $ 8.25 . we ' re asked for the price of the largest ( most expensive ) pot . since the total price is $ 8.25 ( a 15 - cent increment ) and the the difference in sequential prices of the pots is 15 cents , the largest pot probably has a price that is a 15 - cent increment . from the answer choices , i would then test answer a first if . . . . the largest pot = $ 1.75 1 1.15 1.3 1.45 1.6 1.75 total = $ 8.25 so this must be the answer . a" | a = 6 - 1
b = a * 6
c = b / 2
d = c * 0
e = 8 - 25
f = e / 6
g = 6 - 1
h = g * 0
i = f + h
|
a ) 5 min , b ) 2 min , c ) 6 min , d ) 1 min , e ) 10 min | c | multiply(66, divide(1, 11)) | a pipe can fill a cistern in 66 minutes . find the time in which 1 / 11 part of the cistern will be filled ? | "full cistern filled in = 66 minutes 1 / 11 part filled in = 66 * 1 / 11 = 6 minutes answer is c" | a = 1 / 11
b = 66 * a
|
a ) 110 , b ) 887 , c ) 167 , d ) 197 , e ) 161 | a | subtract(multiply(250, divide(15, divide(15, const_3))), multiply(160, divide(20, divide(15, const_3)))) | a train crosses a platform of 160 m in 15 sec , same train crosses another platform of length 250 m in 20 sec . then find the length of the train ? | "length of the train be β x β x + 160 / 15 = x + 250 / 20 20 x + 3200 = 15 x + 3750 5 x = 550 x = 110 m answer : a" | a = 15 / 3
b = 15 / a
c = 250 * b
d = 15 / 3
e = 20 / d
f = 160 * e
g = c - f
|
a ) 9 / 4 , b ) 3 / 2 , c ) 4 / 3 , d ) 2 / 3 , e ) 1 / 2 | b | divide(2, 5) | a positive number x is multiplied by 2 , and this product is then divided by 5 . if the positive square root of the result of these two operations equals x , what is the value of x ? | "we need to produce an equation from the information given in the problem stem . we are first given that x is multiplied by 2 and then the product is divided by 3 . this gives us : 2 x / 3 next we are given that the positive square root of the result ( which is 2 x / 3 ) is equal to x . this gives us β ( 2 x / 3 ) = x 2 x / 3 = x ^ 2 2 x = 3 x ^ 2 3 x ^ 2 β 2 x = 0 x ( 3 x β 2 ) = 0 x = 0 or 3 x β 2 = 0 3 x = 2 x = 3 / 2 because x is positive , x = 3 / 2 . the answer is b ." | a = 2 / 5
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | multiply(multiply(multiply(add(multiply(const_3, const_10), const_1), const_2), const_4), 11) | if the number 892 , 132,26 x is divisible by 11 , what must be the value of x ? | "multiplication rule of 11 : ( sum of digits at odd places - sum of digits at even places ) should be divisible by 11 given number : 892 , 132,24 x sum of digits at odd places = 8 + 2 + 3 + 2 + x = 15 + x ( i ) sum of digits at even places = 9 + 1 + 2 + 6 = 18 ( ii ) ( i ) - ( ii ) = 15 + x - 18 = x - 3 hence x should be = 3 to make this a multiple of 11 ( 0 ) option c" | a = 3 * 10
b = a + 1
c = b * 2
d = c * 4
e = d * 11
|
a ) 48 , b ) 9 , c ) 7 , d ) 66 , e ) 15 | d | subtract(divide(multiply(1.75, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.10, const_1000), divide(multiply(90, const_1000), const_3600))) | two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.75 km and 1.10 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | ": relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.75 + 1.10 = 2.75 km = 2750 m . required time = 2750 * 3 / 125 = 66 sec . answer : d" | a = 1 * 75
b = 60 * 1000
c = b / 3600
d = a / c
e = 1 * 10
f = 90 * 1000
g = f / 3600
h = e / g
i = d - h
|
a ) 39 : 40 , b ) 39 : 45 , c ) 39 : 40 , d ) 39 : 41 , e ) 39 : 42 | a | divide(add(300, multiply(divide(300, add(subtract(400, const_100), 300)), add(210, multiply(const_2, add(const_3, const_2))))), subtract(400, const_100)) | a and b put in rs . 300 and rs . 400 respectively into a business . a reinvests into the business his share of the first year ' s profit of rs . 210 where as b does not . in what ratio should they divide the second year ' s profit ? | "3 : 4 a = 3 / 7 * 210 = 90 390 : 400 39 : 40 answer : a" | a = 400 - 100
b = a + 300
c = 300 / b
d = 3 + 2
e = 2 * d
f = 210 + e
g = c * f
h = 300 + g
i = 400 - 100
j = h / i
|
a ) rs . 169.50 , b ) rs . 1700 , c ) rs . 175.50 , d ) rs . 180 , e ) none | c | divide(subtract(multiply(153, add(add(1, 1), 2)), add(126, 126)), 2) | tea worth rs . 126 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 153 per kg , the price of the third variety per kg will be | "solution since first 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 . x - 153 / 22.50 = 1 = βΊ x - 153 = 22.50 = βΊ x = 175.50 . hence , price of the third variety = rs . 175.50 per kg . answer c" | a = 1 + 1
b = a + 2
c = 153 * b
d = 126 + 126
e = c - d
f = e / 2
|
a ) 40 , b ) 48 , c ) 56 , d ) 62 , e ) 54 | e | multiply(divide(144, power(const_2, 5)), const_2) | if a and b are integers and ( a * b ) ^ 5 = 144 y , y could be : | "distribute the exponent . a ^ 5 * b ^ 5 = 144 y find the prime factorization of 144 . this is 2 ^ 4 * 3 ^ 2 . we need 2 ^ 1 and 3 ^ 3 i . e 2 * 27 = 54 the answer is e ." | a = 2 ** 5
b = 144 / a
c = b * 2
|
a ) 1 , b ) 2 , c ) 4 , d ) 5 , e ) 7 | d | floor(divide(23, 4)) | on dividing 23 by a number , the quotient is 4 and the remainder is 3 . find the divisor . | "d = ( d - r ) / q = ( 23 - 3 ) / 4 = 20 / 4 = 5 d" | a = 23 / 4
b = math.floor(a)
|
['a ) 18 cm', 'b ) 28 cm', 'c ) 30 cm', 'd ) 38 cm', 'e ) 42 cm'] | a | multiply(const_2, sqrt(add(41, multiply(const_2, 20)))) | the diagonal of a rectangle is 41 cm and its area is 20 sq . cm . the perimeter of the rectangle must be : | l 2 + b 2 = 41 . also , lb = 20 . ( l + b ) 2 = ( l 2 + b 2 ) + 2 lb = 41 + 40 = 81 ( l + b ) = 9 . perimeter = 2 ( l + b ) = 18 cm . a ) | a = 2 * 20
b = 41 + a
c = math.sqrt(b)
d = 2 * c
|
a ) 12 km , b ) 13 km , c ) 14 km , d ) 15 km , e ) 16 km | a | divide(add(divide(7, const_60), divide(8, const_60)), divide(const_1, 16)) | a boy is travelling from his home to school at 12 km / hr and reached 7 min late . next day he traveled at 16 km / hr and reached 8 min early . distance between home and school ? | "let the distance be x t 1 = x / 12 hr t 2 = x / 16 hr difference in time = 7 + 8 = 15 = 1 / 4 hr x / 12 - x / 16 = 1 / 4 x / 48 = 1 / 4 x = 12 km answer is a" | a = 7 / const_60
b = 8 / const_60
c = a + b
d = 1 / 16
e = c / d
|
a ) 4 : 1 , b ) 1 : 4 , c ) 3 : 1 , d ) 2 : 3 , e ) 2 : 5 | c | divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), multiply(add(const_2, const_3), const_2))) | if a and b get profits of rs . 15,000 and rs . 5,000 respectively at the end of year then ratio of their investments are | "ratio = 15000 / 5000 = 3 : 1 answer : c" | a = 3 * 2
b = a * 100
c = b * 100
d = 3 * 2
e = d * 100
f = e * 100
g = 2 + 3
h = g * 2
i = f / h
j = c / i
|
['a ) 2.5', 'b ) 3.0', 'c ) 4.0', 'd ) 6.4', 'e ) 8.0'] | d | divide(divide(0.4, const_0_25), const_0_25) | after an ice began to melt out from the freezer , in the first hour lost 3 / 4 , in the second hour lost 3 / 4 of its remaining . if after two hours , the volume is 0.4 cubic inches , what is the original volume of the cubic ice , in cubic inches ? | let initial volume of ice be = x ice remaining after 1 hour = x - 0.75 x = 0.25 x ice remaining after 2 hour = ( 1 / 4 ) x - ( 3 / 4 * 1 / 4 * x ) = ( 1 / 16 ) x ( 1 / 16 ) x = 0.4 x = 6.4 alternate solution : try to backsolve . initial volume = 6.4 after one hour - - > ( 1 / 4 ) 6.4 = 1.6 after two hours - - > ( 1 / 4 ) 1.6 = 0.4 answer : d | a = 0 / 4
b = a / const_0_25
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a ) 79 kmph , b ) 85 kmph , c ) 35 kmph , d ) 23 kmph , e ) 14 kmph | c | divide(add(10, 60), const_2) | the speed of a car is 10 km in the first hour and 60 km in the second hour . what is the average speed of the car ? | "s = ( 10 + 60 ) / 2 = 35 kmph answer : c" | a = 10 + 60
b = a / 2
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a ) 327 , b ) 645 , c ) 987 , d ) 1119 , e ) 1339 | d | subtract(436,600, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 324,700 and 436,600 have tens digit 1 and units digit 3 ? | "the integers are : 324,713 324,813 etc . . . 436,513 the number of integers is 4366 - 3247 = 1119 the answer is d ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 436 - 600
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a ) 54 , b ) 72 , c ) 96 , d ) 144 , e ) 432 | a | multiply(subtract(const_10, 4), 9) | plaudia can choose any two of 4 different candles and any 8 of 9 different flowers for a centerpiece arrangement . given these choices , how many candle + flower groupings can she select ? | plaudia can choose any two of four different candles and any 8 of 9 different flowers for a centerpiece arrangement . 2 candles from 4 candles : 4 c 2 : 6 8 flowers from 9 flowers : 9 c 8 : 9 so total grouping 9 * 6 = 54 . answer is a . | a = 10 - 4
b = a * 9
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a ) 14 % , b ) 25 % , c ) 32 % , d ) 40 % , e ) 52 % | a | multiply(divide(subtract(64, 55), 64), const_100) | in town x , 64 percent of the population are employed , and 55 percent of the population are employed males . what percent of the employed people in town x are females ? | "we are asked to find the percentage of females in employed people . total employed people 64 % , out of which 55 are employed males , hence 9 % are employed females . ( employed females ) / ( total employed people ) = 9 / 64 = 14 % answer : a ." | a = 64 - 55
b = a / 64
c = b * 100
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a ) 12 % , b ) 26 % , c ) 29 % , d ) 32 % , e ) 60 % | d | multiply(divide(add(multiply(divide(30, const_100), 300), multiply(divide(35, const_100), 200)), add(300, 200)), const_100) | for an agricultural experiment , 300 seeds were planted in one plot and 200 were planted in a second plot . if exactly 30 percent of the seeds in the first plot germinated and exactly 35 percent of the seeds in the second plot germinated , what percent of the total number of seeds germinated ? | "in the first plot 30 % of 300 seeds germinated , so 0.3 x 300 = 90 seeds germinated . in the second plot , 35 % of 200 seeds germinated , so 0.35 x 200 = 70 seeds germinated . since 90 + 70 = 260 seeds germinated out of a total of 300 + 200 = 500 seeds , the percent of seeds that germinated is ( 160 / 500 ) x 100 % , or 32 % . answer : d ." | a = 30 / 100
b = a * 300
c = 35 / 100
d = c * 200
e = b + d
f = 300 + 200
g = e / f
h = g * 100
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a ) 110 , b ) 100 , c ) 190 , d ) 200 , e ) 210 | e | subtract(multiply(divide(20, 2), add(divide(20, 2), 1)), multiply(divide(add(19, 1), 2), add(divide(subtract(19, 1), 2), 1))) | ifaequals the sum of the even integers from 2 to 20 , inclusive , andbequals the sum of the odd integers from 1 to 19 , inclusive , what is the value of a + b ? | "answer is 10 yes ! there is really a faster way to solve it . sum of consecutive odd or even integers = ( no . of odd or even ints ) * ( first int + last int ) / 2 here a = sum of even ints from 2 to 20 , inclusive number of even ints = 10 , first int + last int = 2 + 20 = 22 a = 10 * 22 / 2 = 110 b = sum of odd ints from 1 to 19 , inclusive number of odd ints = 10 , first int + last int = 1 + 19 = 20 a = 10 * 20 / 2 = 100 a + b = 110 + 100 = 210 ans : e" | a = 20 / 2
b = 20 / 2
c = b + 1
d = a * c
e = 19 + 1
f = e / 2
g = 19 - 1
h = g / 2
i = h + 1
j = f * i
k = d - j
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a ) 35 kmph , b ) 48 kmph , c ) 52 kmph , d ) 54 kmph , e ) 60 kmph | e | divide(540, divide(multiply(6, 3), 2)) | a car takes 6 hours to cover a distance of 540 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 6 distance = 540 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 540 / 9 = 60 kmph e )" | a = 6 * 3
b = a / 2
c = 540 / b
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a ) 52 sec , b ) 45 sec , c ) 60 sec , d ) 25 sec , e ) 39 sec | c | divide(multiply(50, const_4), multiply(12, divide(const_1000, const_3600))) | how long will a boy take to run round a square field of side 50 meters , if he runs at the rate of 12 km / hr ? | "speed = 12 km / hr = 12 * 5 / 18 = 10 / 3 m / sec distance = 50 * 4 = 200 m time taken = 200 * 3 / 10 = 60 sec answer is c" | a = 50 * 4
b = 1000 / 3600
c = 12 * b
d = a / c
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a ) 28 , b ) 27 , c ) 12 , d ) 37 , e ) 01 | d | add(subtract(85, multiply(17, 3)), 3) | a batsman in his 17 th innings makes a score of 85 and their by increasing his average by 3 . what is his average after the 17 th innings ? | "explanation : 16 x + 85 = 17 ( x + 3 ) x = 34 + 3 = 37 answer : d" | a = 17 * 3
b = 85 - a
c = b + 3
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a ) 900 , b ) 1,800 , c ) 2,700 , d ) 10,800 , e ) 64,800 | a | multiply(multiply(divide(270, 6), 4), 5) | running at the same constant rate , 6 identical machines can produce a total of 270 bottles per minute . at this rate , how many bottles could 5 such machines produce in 4 minutes ? | "6 machines produce 270 bottles per minute ; 1 machine produces 270 / 6 = 45 bottles per minute ; 5 machines produce 45 * 5 = 225 bottles per minute ; in 4 minutes 10 machines produce 225 * 4 = 900 bottles . answer : a ." | a = 270 / 6
b = a * 4
c = b * 5
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a ) 37.5 km , b ) 50.5 km , c ) 60 km , d ) 85.5 km , e ) 95 km | a | multiply(const_60, divide(multiply(divide(15, const_60), 50), 50)) | when a train travels at a speed of 70 kmph , it reaches the destination on time . when the same train travels at a speed of 50 kmph , it reaches its destination 15 min late . what is the length of journey ? | "let x be the time reached with the speed 70 km / h 50 km / h - - - - > x + 15 distance is equal so 70 ( km / h ) Γ xhr = 50 ( km / h ) Γ ( x + 15 ) hr so 70 x = 50 x + 750 so the would be in km and x = 37.5 answer : a" | a = 15 / const_60
b = a * 50
c = b / 50
d = const_60 * c
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a ) 2400 , b ) 3600 , c ) 3998 , d ) 2539 , e ) 1930 | b | divide(multiply(9000, multiply(const_2, const_2)), add(add(multiply(divide(multiply(const_2, const_2), const_3), const_3), multiply(const_1, const_3)), multiply(const_1, const_3))) | p , q and r have $ 9000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? | "b 3600 let the amount with r be $ r r = 2 / 3 ( total amount with p and q ) r = 2 / 3 ( 9000 - r ) = > 3 r = 18000 - 2 r = > 5 r = 18000 = > r = 3600 ." | a = 2 * 2
b = 9000 * a
c = 2 * 2
d = c / 3
e = d * 3
f = 1 * 3
g = e + f
h = 1 * 3
i = g + h
j = b / i
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a ) 1 / 3 , b ) 1 / 2 , c ) 7 / 8 , d ) 3 / 4 , e ) 7 / 3 | b | divide(const_6, multiply(const_6, const_6)) | if two dice are thrown together , the probability of getting an even number on one die and an odd number on the other is - . | "answer : option b the number of exhaustive outcomes is 36 . let e be the event of getting an even number on one die and an odd number on the other . let the event of getting either both even or both odd then = 18 / 36 = 1 / 2 p ( e ) = 1 - 1 / 2 = 1 / 2 ." | a = 6 * 6
b = 6 / a
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a ) 58 , b ) 59 , c ) 64 , d ) 70 , e ) 74 | c | add(add(multiply(divide(60, 2), const_2), 2), add(divide(60, 2), 2)) | the sum of the present ages of two persons a and b is 60 . if the age of a is twice that of b , find the sum of their ages 2 years hence ? | "explanation : a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 2 years , their ages will be 42 and 22 . sum of their ages = 42 + 22 = 64 . c )" | a = 60 / 2
b = a * 2
c = b + 2
d = 60 / 2
e = d + 2
f = c + e
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a ) 1 , b ) 3 , c ) 4 , d ) 8 , e ) 9 | d | add(add(const_4, const_3), const_2) | what is the units digit of the expression 13 ^ 27 + 1 ? | "power of 3 follows repetition pattern of = 3 , 9 , 7 , 1 , 3 , 9 , 7 , 1 . . . . repetition after 4 powers = ( 3 , 9 , 7 , 1 ) 13 ^ 27 = 27 th power = 24 + 3 24 / 4 = 6 ( 6 repetitions ) + 3 = ` ` 7 ' ' 7 + 1 = 8 answer d" | a = 4 + 3
b = a + 2
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a ) 18.9 sec , b ) 88.9 sec , c ) 22.9 sec , d ) 27.00 sec , e ) 72.0 sec | d | divide(add(110, 340), multiply(60, const_0_2778)) | how long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 340 m in length ? | "d = 110 + 340 = 450 m s = 60 * 5 / 18 = 50 / 3 t = 450 * 3 / 50 = 27.00 sec answer : d" | a = 110 + 340
b = 60 * const_0_2778
c = a / b
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a ) 18 , b ) 20 , c ) 22 , d ) 24 , e ) 26 | d | divide(multiply(30, const_4), add(const_4, const_1)) | the cost price of 30 articles is the same as the selling price of x articles . if the profit is 25 % , what is x ? | let the cost price = y the cost price of 30 articles = 30 y the selling price of x articles = 1.25 y * x 1.25 y * x = 30 y x = 30 / 1.25 = 24 the answer is d . | a = 30 * 4
b = 4 + 1
c = a / b
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a ) 16 , b ) 9 , c ) 8 , d ) 11 , e ) 10 | a | divide(multiply(subtract(42, const_2), const_2), add(const_4, const_1)) | a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 42 , the how old is b ? | "explanation : let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 42 β 5 x = 40 β x = 8 . hence , b ' s age = 2 x = 16 years . answer : a" | a = 42 - 2
b = a * 2
c = 4 + 1
d = b / c
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a ) 3 : 5 , b ) 3 : 7 , c ) 3 : 8 , d ) 3 : 10 , e ) 3 : 9 | d | divide(15, 50) | calculate the ratio between x and y if 15 % of x equal to 50 % of y ? | "explanation : 15 x = 50 y x : y = 15 : 50 = 3 : 10 answer : d" | a = 15 / 50
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a ) 2 : 3 , b ) 3 : 4 , c ) 9 : 80 , d ) 20 : 3 , e ) none | c | divide(multiply(0.09, const_100), multiply(0.8, const_100)) | if 0.8 of a number is equal to 0.09 of another number , the ratio of the numbers i | "sol . 0.8 a = 0.09 b Γ’ β‘ β a / b = 0.09 / 0.8 = 9 / 80 = . Γ’ Λ Β΄ a : b = 9 : 80 . answer c" | a = 0 * 9
b = 0 * 8
c = a / b
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