regisss/math_qa huggingface dataset:
Problem: if x is 30 percent more than y and y is 40 percent less than z , then x is what percent of z ?
Options: a ) 78 % , b ) 250 % , c ) 500 / 3 % , d ) 125 % , e ) 60 %
Rationale: "z = 100 ; y = 60 so x = 78 x as % of z = 78 / 100 * 100 = > 78 % answer will be ( a )"
Answer: a
Problem: a train running at the speed of 60 km / hr crosses a pole in 12 seconds . find the length of the train .
Options: a ) 150 , b ) 278 , c ) 179 , d ) 200 , e ) 191
Rationale: ": speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 12 = 200 meter answer : d"
Answer: d
Problem: if y > 0 , ( 7 y ) / 20 + ( 3 y ) / 10 is what percent of y ?
Options: a ) 40 % , b ) 50 % , c ) 65 % , d ) 70 % , e ) 80 %
Rationale: "can be reduced to 7 y / 20 + 6 y / 20 = 13 y / 20 = 65 % answer c"
Answer: c
Problem: 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 ?
Options: a ) $ 257615 , b ) $ 437615 , c ) $ 127615 , d ) $ 117615 , e ) $ 157615
Rationale: "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"
Answer: d
Problem: what is the cp of rs 100 stock at 9 discount , with 1 / 5 % brokerage ?
Options: a ) 96.9 , b ) 96.3 , c ) 91.2 , d ) 96.7 , e ) 96.21
Rationale: "explanation : use the formula , cp = 100 â € “ discount + brokerage % cp = 100 - 9 + 1 / 5 91.2 thus the cp is rs 91.2 . answer : c"
Answer: c
Problem: 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 ?
Options: a ) 500 , b ) 1,600 / 3 , c ) 1,000 , d ) 1,500 , e ) 2,000
Rationale: 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
Answer: e
Problem: 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 ?
Options: a ) 51 , b ) 87 , c ) 35 , d ) 75 , e ) 68
Rationale: 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
Answer: c
Problem: 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
Options: a ) 110 m , b ) 120 m , c ) 140 m , d ) 160 m , e ) 100 cm
Rationale: "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"
Answer: e
Problem: 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 ?
Options: a ) 4038 , b ) 8076 , c ) 9691.2 , d ) 4845.6 , e ) none of these
Rationale: ∵ amount , he have spent in 1 month on clothes transport = amount spent on saving per month ∵ amount , spent on clothes and transport = 48456 ⁄ 12 = 4038 answer a
Answer: a
Problem: what is the sum of all the multiples of 10 between 0 and 100 ?
Options: a ) 500 , b ) 620 , c ) 550 , d ) 340 , e ) 440
Rationale: "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"
Answer: c
Problem: 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 ?
Options: a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90
Rationale: "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"
Answer: c
Problem: difference of two numbers is 1650 . if 7.5 % of the number is 12.5 % of the other number , find the number ?
Options: a ) 2660 , 1000 , b ) 3660 , 2000 , c ) 3000 , 4160 , d ) 2490 , 4150 , e ) 2475 , 4125
Rationale: "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 – y = 1650 2 * y / 3 = 1650 y = [ ( 1650 * 3 ) / 2 ] = 2475 . one number = 2475 , second number = 5 * y / 3 = 4125 . answer e ."
Answer: e
Problem: a certain no . when divided by 140 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ?
Options: a ) 2 , b ) 4 , c ) 7 , d ) 8 , e ) 11
Rationale: "explanation : 140 + 25 = 165 / 15 = 11 ( remainder ) e"
Answer: e
Problem: 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 .
Options: a ) 12 , b ) 200 , c ) 600 , d ) 800 , e ) 20
Rationale: "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"
Answer: a
Problem: an inspector rejects 0.07 % of the meters as defective . how many will he examine to reject 2 ?
Options: a ) a ) 1500 , b ) b ) 2000 , c ) c ) 2857 , d ) d ) 3000 , e ) e ) 3100
Rationale: "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"
Answer: c
Problem: 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 ?
Options: a ) rs . 147.50 , b ) rs . 785.50 , c ) rs . 176.50 , d ) rs . 258.50 , e ) none of these
Rationale: "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 â ˆ ’ 154 22.50 = > x â ˆ ’ ( 154 / 22.50 ) = 1 . = > x â ˆ ’ 154 = 22.50 . = > x = 176.50 rs . answer : c"
Answer: c
Problem: 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 ?
Options: a ) 46 , b ) 47 , c ) 48 , d ) 49 , e ) 50
Rationale: "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"
Answer: e
Problem: in a simultaneous throw of a pair of dice , find the probability of getting a total more than 10
Options: a ) 5 / 13 , b ) 2 / 15 , c ) 1 / 12 , d ) 3 / 17 , e ) 6 / 19
Rationale: "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"
Answer: c
Problem: 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 ?
Options: a ) 5.5 kg , b ) 6 kg , c ) 30 kg , d ) 36.5 kg , e ) 71 kg
Rationale: 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
Answer: b
Problem: 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
Options: a ) 10 , b ) 12 , c ) 16 , d ) 22 , e ) 20
Rationale: sol . let the numbers be x and ( 44 - x ) . then , 5 x = 6 ( 44 - x ) ⇔ 11 x = 264 ⇔ x = 24 . so , the numbers are 24 and 20 . answer e
Answer: e
Problem: 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 ?
Options: a ) 4 , b ) 6 , c ) 10 , d ) 15 , e ) 20
Rationale: "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"
Answer: b
Problem: 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 ?
Options: a ) 19 , b ) 12 , c ) 11 , d ) 30 , e ) 114
Rationale: 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
Answer: d
Problem: 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 )
Options: a ) 1
b ) 2
c ) 3
d ) 4
e ) 5
Rationale: 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 .
Answer: e
Problem: 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 ?
Options: a ) 244 ft , b ) 88 ft , c ) 122 ft , d ) 178 ft , e ) 66 ft
Rationale: "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"
Answer: d
Problem: a train 400 m in length crosses a telegraph post in 16 seconds . the speed of the train is ?
Options: a ) 22 kmph , b ) 77 kmph , c ) 54 kmph , d ) 71 kmph , e ) 90 kmph
Rationale: "s = 400 / 16 * 18 / 5 = 90 kmph answer : e"
Answer: e
Problem: ravi can do a piece of work in 50 days while prakash can do it in 75 days . in how many days will they finish it together ?
Options: a ) 30 days , b ) 35 days , c ) 25 days , d ) 27 days , e ) 29 days
Rationale: 1 / 50 + 1 / 75 = 5 / 150 30 / 1 = 30 days answer : a
Answer: a
Problem: the marks obtained by polly and sandy are in the ratio 3 : 5 and those obtained by sandy and willy are in the ratio of 5 : 2 . the marks obtained by polly and willy are in the ratio of . . . ?
Options: a ) 2 : 1 , b ) 3 : 2 , c ) 4 : 3 , d ) 5 : 4 , e ) 6 : 5
Rationale: "polly : sandy = 3 : 5 sandy : willy = 5 : 2 polly : sandy : willy = 3 : 5 : 2 polly : willy = 3 : 2 the answer is b ."
Answer: b
Problem: add 10 % of 30 and 15 % of 50 .
Options: a ) 9.5 , b ) 10.5 , c ) 11.5 , d ) 12 , e ) 15
Rationale: "10 % of 30 + 15 % of 50 30 * 10 / 100 + 50 * 15 / 100 3 + 7.5 = 10.5 answer b"
Answer: b
Problem: smita was making a cube with dimensions 5 * 5 * 5 using 1 * 1 * 1 cubes . what is the number of cubes needed to make a hollow cube looking of the same shape ?
Options: a ) 98
b ) 104
c ) 100
d ) 61
e ) 51
Rationale: cube use to make perfect 5 * 5 * 5 = 125 no need of cube to make hollow = ( 5 - 2 ) * ( 5 - 2 ) * ( 5 - 2 ) = 27 so total cube need to make hollow cube is = 125 - 27 = 98 answer : a
Answer: a
Problem: the average of temperatures at noontime from monday to friday is 50 ; the lowest one is 40 , what is the possible maximum range of the temperatures ?
Options: a ) 20 , b ) 25 , c ) 40 , d ) 50 , e ) 75
Rationale: "there are 5 days so the sum of temperature can be 50 * 5 = 250 lowest is 40 . to find the maximum range we can say the temperature was the lowest for 4 of the 5 days so 4 * 40 = 160 . on the fifth day it is 250 - 160 = 90 range is therefore 90 - 40 = 50 answer d"
Answer: d
Problem: 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 30 paisa . if the share of y is rs . 36 , what is the total amount ?
Options: a ) 166 , b ) 140 , c ) 178 , d ) 177 , e ) 169
Rationale: "x : y : z = 100 : 45 : 30 20 : 9 : 6 9 - - - 36 35 - - - ? = > 140 answer : b"
Answer: b
Problem: a retailer sold 2 houses at each $ 10000 . on 1 st house he got 30 % profit and on 2 nd house he loss 10 % then find his net profit or loss ?
Options: a ) 15 % , b ) 17 % , c ) 21 % , d ) 19 % , e ) 25 %
Rationale: a = 30 % b = - 10 % profit % = 30 - 10 + ( - 300 / 100 ) = 17 % answer is b
Answer: b
Problem: the remainder when e = 1 + 3 + 3 ^ 2 + 3 ^ 3 + . . . . . . . . . . + 3 ^ 200 is divided 13 .
Options: a ) 12 , b ) 7 , c ) 0 , d ) 5 , e ) 3
Rationale: e = 1 + 3 + 3 ^ 2 + 3 ^ 3 + . . . . . . . . . . . . . . . . . . + 3 ^ 200 is a geometric progression having common ratio as ' 3 ' and number of terms as ' 201 ' . since sum to n terms in gp = a ( r ^ n - 1 ) / ( r - 1 ) where a = first term and r = common ration hence , 1 * ( 3 ^ 201 - 1 ) / ( 3 - 1 ) rem of ( 3 ^ 201 - 1 ) / 2 divided by 13 3 ^ 201 - 1 / 26 wkt , 3 ^ 3 = 27 = 26 + 1 { ( 26 + 1 ) ^ 67 - 1 } / 26 { 1 - 1 } / 26 = > 0 . c
Answer: c
Problem: today joelle opened an interest - bearing savings account and deposited $ 6000 . if the annual interest rate is 5 percent compounded interest , and she neither deposits nor withdraws money for exactly 2 years , how much money will she have in the account ?
Options: a ) $ 6715 , b ) $ 5615 , c ) $ 6415 , d ) $ 6615 , e ) $ 6315
Rationale: interest for 1 st year = 6000 * 5 / 100 = 300 interest for 2 nd year = 6300 * 5 / 100 = 315 total = 6000 + 300 + 315 = 6615 answer : d
Answer: d
Problem: a boat can travel with a speed of 24 km / hr in still water . if the speed of the stream is 4 km / hr , find the time taken by the boat to go 196 km downstream
Options: a ) 3 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours
Rationale: "explanation : speed of the boat in still water = 24 km / hr speed of the stream = 4 km / hr speed downstream = ( 22 + 5 ) = 28 km / hr distance travelled downstream = 196 km time taken = distance / speed = 196 / 28 = 7 hours answer : option e"
Answer: e
Problem: a train 900 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 ?
Options: a ) 200 m , b ) 379 m , c ) 400 m , d ) 589 m , e ) 600 m
Rationale: "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 900 + x ) / 60 = 65 / 3 x = 400 m . answer : c"
Answer: c
Problem: jayant opened a shop investing rs . 30,000 . madhu joined him 2 months later , investing rs . 45,000 . they earned a profit of rs . 54,000 after completion of one year . what will be madhu ' s share of profit ?
Options: a ) rs . 27,000 , b ) rs . 24,000 , c ) rs . 30,000 , d ) rs . 36,000 , e ) none of these
Rationale: "30,000 * 12 = 45,000 * 8 1 : 1 madhu ' s share = 1 / 2 * 54,000 i . e . rs . 27,000 answer : a"
Answer: a
Problem: 600 men have provisions for 20 days . if 200 more men join them , for how many days will the provisions last now ?
Options: a ) 15 , b ) 11 , c ) 10 , d ) 8 , e ) 7
Rationale: "600 * 20 = 800 * x x = 15 . answer : a"
Answer: a
Problem: the hcf of two numbers is 52 and the other two factors of their lcm are 11 and 12 . what is the largest number .
Options: a ) 624 , b ) 450 , c ) 480 , d ) 504 , e ) 555
Rationale: "explanation : hcf of the two numbers = 52 hcf will be always a factor of lcm 52 is factor of lcm other two factors are 11 & 12 then the numbers are ( 52 * 11 ) and ( 52 x 12 ) = 572 and 624 answer : option a"
Answer: a
Problem: a whale goes on a feeding frenzy that lasts for 9 hours . for the first hour he catches and eats x kilos of plankton . in every hour after the first , it consumes 3 kilos of plankton more than it consumed in the previous hour . if by the end of the frenzy the whale will have consumed a whopping accumulated total 270 kilos of plankton , how many kilos did he consume on the sixth hour ?
Options: a ) 33 , b ) 47 , c ) 50 , d ) 53 , e ) 62
Rationale: "if you list the amount eaten each hour , you ' ll get an equally spaced list , increasing by 3 each hour . in any equally spaced list , the median equals the mean . here , the mean is 270 / 9 = 30 , so the median is also 30 , and that is the amount eaten in the 5 th hour . we need to add 3 to find the total eaten in the next hour , so the answer is 33 . option a"
Answer: a
Problem: in a garment industry , 12 men working 8 hours per day complete a piece of work in 10 days . to complete the same work in 8 days , working 5 hours a day , the number of men required is :
Options: a ) 24 , b ) 5 , c ) 6 , d ) 8 , e ) 9
Rationale: "explanation : let the required number of men be x . less days , more men ( indirect proportion ) more working hrs per day , less men ( indirect proportion ) days 8 : 10 working hrs 5 : 8 : : 12 : x = > 8 x 5 x x = 10 x 8 x 12 = > x = 10 x 8 x 12 / ( 8 x 5 ) = > x = 24 answer : a"
Answer: a
Problem: if the radii of umbra and penumbra cast by an object on a wall are of the ratio 2 : 6 , what is the area of the penumbra ring around the umbra of the latter ’ s radius is 40 cms ?
Options: a ) 40288.57 cm ^ 2
b ) 40388.57 cm ^ 2
c ) 40488.57 cm ^ 2
d ) 40588.57 cm ^ 2
e ) 40688.57 cm ^ 2
Rationale: et the radius of umbra and penumbra are 2 k and 6 k . then as given radius of umbra = 40 cm so 2 k = 40 k = 20 radius of penumbra = 20 * 6 = 120 area of penumbra ring around the umbra = area of penumbra - area of umbra 22 / 7 * [ ( 120 ) ^ 2 - ( 40 ) ^ 2 ] = 40288.57 cm ^ 2 answer : a
Answer: a
Problem: if in a box of dimensions 6 m * 5 m * 4 m smaller boxes of dimensions 60 cm * 50 cm * 40 cm are kept in it , then what will be the maximum number of the small boxes that can be kept in it ?
Options: a ) 1098 , b ) 1000 , c ) 1628 , d ) 1098 , e ) 1094
Rationale: 6 * 5 * 4 = 60 / 100 * 50 / 100 * 40 / 100 * x 1 = 1 / 10 * 1 / 10 * 1 / 10 * x = > x = 1000 answer : b
Answer: b
Problem: the area of a square field is 64816 sq m . how long will a lady take to cross the field diagonally at the rate of 7.2 km / hr ?
Options: a ) 2 min , b ) 4 min , c ) 5 min , d ) 3 min , e ) 3.5 min
Rationale: "area of a square field = 64816 sq m let the side of square = a a ^ 2 = 64816 = > a = 254.59 diagonal = ( 2 ) ^ ( 1 / 2 ) * a = 1.414 * 254.59 = 360 speed of lady = 7.2 km / hour = 7200 m / hour = 120 m / min time taken by lady to cross the field diagonally = 360 / 120 = 3 min answer d"
Answer: d
Problem: the ratio 15 : 25 expressed as percent equals to
Options: a ) 70 % , b ) 60 % , c ) 30 % , d ) 75 % , e ) none of above
Rationale: "explanation : actually it means 15 is what percent of 25 , which can be calculated as , ( 15 / 25 ) * 100 = 15 * 4 = 60 answer : option b"
Answer: b
Problem: the ratio between the number of sheep and the number of horses at the stewar farm is 2 to 7 . if each of horse is fed 230 ounces of horse food per day and the farm needs a total 12880 ounces of horse food per day . what is number sheep in the form ? ?
Options: a ) 18 , b ) 16 , c ) 32 , d ) 56 , e ) 58
Rationale: "et no of sheep and horses are 2 k and 7 k no of horses = 12880 / 230 = 56 now 7 k = 56 and k = 8 no of sheep = ( 2 * 8 ) = 16 answer : b"
Answer: b
Problem: the ratio of the arithmetic mean of two numbers to one of the numbers is 4 : 6 . what is the ratio of the smaller number to the larger number ?
Options: a ) 1 : 5 , b ) 1 : 4 , c ) 1 : 3 , d ) 1 : 2 , e ) 2 : 3
Rationale: for two numbers , the arithmetic mean is the middle of the two numbers . the ratio of the mean to the larger number is 4 : 6 , thus the smaller number must have a ratio of 2 . the ratio of the smaller number to the larger number is 2 : 6 = 1 : 3 . the answer is c .
Answer: c
Problem: the speed of a boat in still water in 22 km / hr and the rate of current is 4 km / hr . the distance travelled downstream in 24 minutes is :
Options: a ) 10.6 , b ) 10.2 , c ) 10.8 , d ) 10.4 , e ) 10.0
Rationale: "speed downstream = ( 22 + 4 ) = 26 kmph time = 24 minutes = 24 / 60 hour = 2 / 5 hour distance travelled = time × speed = 2 / 5 × 26 = 10.4 km answer is d ."
Answer: d
Problem: the present ratio of students to teachers at a certain school is 45 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 ?
Options: a ) 4 , b ) 8 , c ) 10 , d ) 12 , e ) 15
Rationale: "we are given that the ratio of students to teacher is 45 to 1 . we can rewrite this using variable multipliers . students : teachers = 45 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  25 / 1 = ( 45 x + 50 ) / ( x + 5 ) if we cross multiply we have : 25 ( x + 5 ) = 45 x + 50 25 x + 125 = 45 x + 50 3.75 = x x ~ 4 since x is the present number of teachers , currently there are 4 teachers . answer a ."
Answer: a
Problem: the number of stamps that kaye and alberto had were in the ration of 5 : 3 respectively . after kaye gave alberto 10 of her stamps , the ration of the number of kaye had to the number of alberto had was 3 : 2 . as a result of the gift , kaye had how many more stamps than alberto ?
Options: a ) 20 , b ) 30 , c ) 40 , d ) 60 , e ) 80
Rationale: "c k 1 = 5 x a 1 = 3 x k 2 = 5 x - 10 a 2 = 3 x + 10 k 2 / a 2 = 3 / 2 ( 5 x - 10 ) / ( 3 x + 10 ) = 3 / 2 2 * ( 5 x - 10 ) = 3 * ( 3 x + 10 ) 10 x - 20 = 9 x + 30 x = 50 k 2 = 5 * 50 - 10 = 240 a 2 = 3 * 50 + 10 = 160 k 2 - a 2 = 80 answer : e"
Answer: e
Problem: a certain bacteria colony doubles in size every day for 16 days , at which point it reaches the limit of its habitat and can no longer grow . if two bacteria colonies start growing simultaneously , how many days will it take them to reach the habitat ’ s limit ?
Options: a ) 6.33 , b ) 7.5 , c ) 10 , d ) 15 , e ) 19
Rationale: if there is one bacteria colony , then it will reach the limit of its habitat in 16 days . if there are two bacteria colonies , then in order to reach the limit of habitat they would need to double one time less than in case with one colony . thus colonies need to double 15 times . answer : d . similar questions to practice : hope it helps .
Answer: d
Problem: find the area of the quadrilateral of one of its diagonals is 15 cm and its off sets 6 cm and 4 cm ?
Options: a ) 50 cm 2 , b ) 60 cm 2 , c ) 75 cm 2 , d ) 90 cm 2 , e ) 150 cm 2
Rationale: "1 / 2 * 15 ( 6 + 4 ) = 75 cm 2 answer : c"
Answer: c
Problem: the average of 10 numbers is 23 . if each number is increased by 1 , what will the new average be ?
Options: a ) 36 , b ) 24 , c ) 72 , d ) 29 , e ) 22
Rationale: "sum of the 10 numbers = 230 if each number is increased by 1 , the total increase = 1 * 10 = 10 the new sum = 230 + 10 = 240 the new average = 240 / 10 = 24 . answer : b"
Answer: b
Problem: sakshi can do a piece of work in 15 days . tanya is 25 % more efficient than sakshi . the number of days taken by tanya to do the same piece of work :
Options: a ) 15 , b ) 16 , c ) 18 , d ) 12 , e ) 10
Rationale: "solution ratio of times taken by sakshi and tanya = 125 : 100 = 5 : 4 . suppose tanya taken x days to do the work . 5 : 4 : : 15 : x ⇒ x = ( 15 x 4 / 5 ) ⇒ x = 12 days . hence , tanya takes 16 days is complete the work . answer d"
Answer: d
Problem: an electric pump can fill a tank in 5 hours . because of a leak in the tank , it took 10 hours to fill the tank . if the tank is full , how much time will the leak take to empty it ?
Options: a ) 10 hours , b ) 12 hours , c ) 8 hours , d ) 5 hours , e ) 15 hours
Rationale: "work done by the leak in 1 hour = 1 / 5 - 1 / 10 = 1 / 10 the leak will empty the tank in 10 hours answer is a"
Answer: a
Problem: 4 , 6 , 12 , 14 , 28 , 30 , . . ?
Options: a ) 20 , b ) 40 , c ) no = 60 , d ) 80 , e ) 100
Rationale: "c 60 the given sequence is a combination of two series 4 , 12 , 28 , . . . . and 6 , 14 , 30 , . . . . the pattern is + 8 , + 16 , + 32 . so , the missing number = ( 28 + 32 ) = 60"
Answer: c
Problem: a cubical block of metal weighs 6 pounds . how much will another cube of the same metal weigh if its sides are twice as long ?
Options: a ) 48 , b ) 99 , c ) 266 , d ) 28 , e ) 17
Rationale: "explanation : if you double the sides of a cube , the ratio of the surface areas of the old and new cubes will be 1 : 4 . the ratio of the volumes of the old and new cubes will be 1 : 8 . weight is proportional to volume . so , if the first weighs 6 pounds , the second weighs 6 x 8 pounds = 48 . answer : a ) 48"
Answer: a
Problem: working together at their respective constant rates , machine a and machine b can produce 1,600 units in 8 hours . working alone , machine b would complete that same output in 50 % more time . if machine a were to work on its own for an 8 - hour shift , what percent of the 1,600 unit total would it produce ?
Options: a ) 25 , b ) 33 , c ) 50 , d ) 67 , e ) 75
Rationale: 1 / a + 1 / b = 1 / t 1 / a + 1 / 12 = 1 / 8 ( 50 % more of 8 is 12 ) 1 / a = 1 / 24 machine a can produce 1600 units in 24 hrs , so it can produce 1600 * 8 / 24 = 533.33 units is 8 hrs . 533.33 is 33.33 % of 1600 . b is the answer
Answer: b
Problem: find the value for x from below equation : x / 3 = - 4 ?
Options: a ) - 12 , b ) 1 , c ) - 2 , d ) - 3 , e ) 4
Rationale: "1 . multiply both sides by 3 : x * 3 / 3 = - 4 / 3 . simplify both sides : x = - 12 a"
Answer: a
Problem: a sum of money is sufficient to pay x ' s wages for 36 days and y ' s wages for 45 days . the same money is sufficient to pay the wages of both for ?
Options: a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20
Rationale: let the total money be $ x x ' s 1 day work = $ x / 36 y ' s 1 day work = $ x / 45 x + y 1 day work = $ x / 20 money is sufficient to pay the wages of both for 20 days answer is e
Answer: e
Problem: length of a rectangular plot is 16 mtr more than its breadth . if the cost of fencing the plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ?
Options: a ) 58 m , b ) 60 m , c ) 80 m , d ) 82 m , e ) 84 m
Rationale: "let breadth = x metres . then , length = ( x + 16 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 16 ) + x ] = 200 2 x + 16 = 100 2 x = 84 x = 42 . hence , length = x + 16 = 58 m a"
Answer: a
Problem: at what price must an article costing rs . 75 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ?
Options: a ) 62.5 , b ) 62.0 , c ) 62.6 , d ) 62.1 , e ) 98.68
Rationale: "cp = 75 sp = 75 * ( 125 / 100 ) = 93.75 mp * ( 95 / 100 ) = 93.75 mp = 98.68 answer : e"
Answer: e
Problem: two stations a and b are 155 km apart on a straight line . one train starts from a at 7 a . m . and travels towards b at 20 kmph . another train starts from b at 8 a . m . and travels towards a at a speed of 25 kmph . at what time will they meet ?
Options: a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 8
Rationale: "suppose they meet x hours after 7 a . m . distance covered by a in x hours = 20 x km . distance covered by b in ( x - 1 ) hours = 25 ( x - 1 ) km . therefore 20 x + 25 ( x - 1 ) = 155 45 x = 180 x = 4 . so , they meet at 11 a . m . answer : option c"
Answer: c
Problem: each week , harry is paid x dollars per hour for the first 12 hours and 1.5 x dollars for each additional hour worked that week . each week , james is paid x dollars per per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week james worked a total of 41 hours if harry and james were paid the same amount last week , how many hours did harry work last week ?
Options: a ) 32 , b ) 36 , c ) 37 , d ) 38 , e ) 39
Rationale: "42 x = 12 x + 1.5 x ( h - 12 ) = = > 42 = 12 + 1.5 ( h - 12 ) = = > h - 12 = 30 / 1.5 = 20 = = > h = 32 answer is a"
Answer: a
Problem: if the volume of two cubes are in the ratio 64 : 1 , the ratio of their edges is :
Options: a ) 3 : 1 , b ) 3 : 2 , c ) 3 : 5 , d ) 4 : 1 , e ) none of these
Rationale: "explanation : let the edges be a and b of two cubes , then a 3 / b 3 = 64 / 1 = > ( a / b ) 3 = ( 4 / 1 ) 3 a / b = 4 / 1 = > a : b = 4 : 1 option d"
Answer: d
Problem: a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 5 : 3 and b : c = 5 : 2 . if the total runs scored by all of them are 60 , the runs scored by b are ?
Options: a ) 20.23 , b ) 20.13 , c ) 30.93 , d ) 19.56 , e ) 10.93
Rationale: "a : b = 5 : 3 b : c = 5 : 2 a : b : c = 25 : 15 : 6 15 / 46 * 60 = 19.56 answer : d"
Answer: d
Problem: a van takes 6 hours to cover a distance of 180 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ?
Options: a ) 50 kmph , b ) 60 kmph , c ) 70 kmph , d ) 80 kmph , e ) 20 kmph
Rationale: "time = 6 distence = 180 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 180 / 9 = 20 kmph e"
Answer: e
Problem: find the principal which yields a simple interest of rs . 20 and compound interest of rs . 30 in two years , at the same percent rate per annum ?
Options: a ) s . 20 , b ) s . 48 , c ) s . 42 , d ) s . 20 , e ) s . 60
Rationale: "explanation : si in 2 years = rs . 20 , si in 1 year = rs . 10 ci in 2 years = rs . 30 % rate per annum = [ ( ci – si ) / ( si in 1 year ) ] * 100 = [ ( 30 – 20 ) / 20 ] * 100 = 50 % p . a . let the principal be rs . x time = t = 2 years % rate = 50 % p . a . si = ( prt / 100 ) 20 = ( x * 50 * 2 ) / 100 x = rs . 20 answer : a"
Answer: a
Problem: a can do a work in 40 days , b in 60 days . a alone start the work after 10 days he left work . in how many days b can finish the work ?
Options: a ) 20 days , b ) 15 days , c ) 30 days , d ) 45 days , e ) 50 days
Rationale: 10 days work of a = 10 / 40 = 1 / 4 remaining work = 1 - 1 / 4 = 3 / 4 b can finish 3 / 4 work = 60 * 3 / 4 = 45 days answer is d
Answer: d
Problem: a certain university will select 1 of 4 candidates eligible to fill a position in the mathematics department and 2 of 7 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ?
Options: a ) 81 , b ) 82 , c ) 84 , d ) 86 , e ) 88
Rationale: 1 c 4 * 2 c 7 = 4 * 21 = 84 the answer is ( c )
Answer: c
Problem: each factor of 150 is inscribed on its own plastic ball , and all of the balls are placed in a jar . if a ball is randomly selected from the jar , what is the probability that the ball is inscribed with a multiple of 15 ?
Options: a ) 1 / 16 , b ) 5 / 42 , c ) 1 / 8 , d ) 3 / 16 , e ) 1 / 3
Rationale: "the # of factors 150 has is 12 so out of 12 factors only four are multiples of 15 : 15 , 30,75 and 150 , itself ; so , the probability is 4 / 12 = 1 / 3 . answer : c ."
Answer: c
Problem: 15.03 ã — 0.01 = ?
Options: a ) 0.1503 , b ) 0.001503 , c ) 1.503 , d ) 0.001503 , e ) none of these
Rationale: "15.03 ã — 0.01 = ? or , ? = 0.1503 answer a"
Answer: a
Problem: 12 men can complete a piece of work in 4 days , while 15 women can complete the same work in 4 days . 6 men start working on the job and after working for 2 days , all of them stopped working . how many women should be put on the job to complete the remaining work , if it is to be completed in 3 days .
Options: a ) 15 , b ) 18 , c ) 22 , d ) 25 , e ) 26
Rationale: one man ’ s one day ’ s work = 1 / 48 one woman ’ s one day ’ s work = 1 / 60 6 men ’ s 2 day ’ s work = ( ( 6 / 48 ) x 2 ) = 1 / 4 remaining work = 3 / 4 now , 1 / 60 work s done in 1 day by 1 woman . thus , 3 / 4 work will be done in 3 days by ( 60 x ( 3 / 4 ) x ( 1 / 3 ) ) = 15 women . answer : a
Answer: a
Problem: mr . kramer , the losing candidate in a two - candidate election , received 942,568 votes , which was exactly 20 percent of all votes cast . approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast ?
Options: a ) 10 % , b ) 12 % , c ) 15 % , d ) 27 % , e ) 37 %
Rationale: "lets assume that candidate got 20 % votes and total votes is 100 . candidate won = 20 remaining = 80 to get 50 % , candidate requires 30 votes from 100 which is 30 % and 30 votes from 80 . 30 / 80 = 37.5 % which is approx 37 % . hence the answer is e"
Answer: e
Problem: in a mixture 60 litres , the ratio of milk and water 2 : 1 . if this ratio is to be 1 : 2 , then the quanity of water to be further added is :
Options: a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60
Rationale: quantity of milk = ( 60 x 2 / 3 ) litres = 40 litres . quantity of water in it = ( 60 - 40 ) litres = 20 litres . new ratio = 1 : 2 let quantity of water to be added further be x litres . then , milk : water = ( 40 / ( 20 + x ) ) now , ( 40 / ( 20 + x ) ) = 1 / 2 = 20 + x = 80 = x = 60 . therefore , quantity of water to be added = 60 litres . answer is e .
Answer: e
Problem: a tap can fill a tank in 16 hours . after half the tank is filled , 3 more similar taps are opened . what is the total time taken to fill the tank completely ?
Options: a ) 2 hrs , b ) 1 hrs , c ) 10 hrs , d ) 5 hrs , e ) 4 hrs
Rationale: explanation : a tap can fill a tank in 16 hours . therefore the tap can fill half the tank in 8 hours . one tap take time to fill half of the tank = 8 hrs 4 taps take time to fill half of the tank = 8 / 4 hrs = 2 hrs total time taken to fill the tank = 8 hr + 2 hr = 10 hrs answer : c
Answer: c
Problem: the sale price sarees listed for rs . 600 after successive discount is 20 % and 5 % is ?
Options: a ) 227 , b ) 358 , c ) 456 , d ) 787 , e ) 191
Rationale: "600 * ( 80 / 100 ) * ( 95 / 100 ) = 456 answer : c"
Answer: c
Problem: a reduction of 25 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 700 , what is the reduced price for kg ?
Options: a ) 35 , b ) 60 , c ) 70 , d ) 80 , e ) 20
Rationale: "a 700 * ( 25 / 100 ) = 175 - - - - 5 ? - - - - 1 = > rs . 35"
Answer: a
Problem: robert is travelling on his cycle and has calculated to reach point a at 2 p . m . if he travels at 10 kmph , he will reach there at 12 noon if he travels at 15 kmph . at what speed must he travel to reach a at 1 p . m . ?
Options: a ) 8 , b ) 12 , c ) 11 , d ) 16 , e ) 18
Rationale: "explanation : let the distance travelled by x km . then , = > ( x / 10 ) - ( x / 15 ) = 2 = > 3 x - 2 x = 60 = > x = 60 = > time taken to travel 60 km at 10 km / hr = 60 / 10 = 6 hours . = > so , robert started 6 hours before 2 p . m . i . e . , at 8 a . m . therefore , speed required = ( 60 / 5 ) km / hr = 12 km / hr answer : b"
Answer: b
Problem: the average of 10 consecutive integers is 16 . then , 9 is deducted from the first consecutive number , 8 is deducted from the second , 7 is deducted form the third , and so on until the last number which remains unchanged . what is the new average ?
Options: a ) 10.5 , b ) 11 , c ) 11.5 , d ) 12 , e ) 12.5
Rationale: the total subtracted is ( 9 + 8 + . . . + 1 ) = ( 9 * 10 ) / 2 = 45 on average , each number will be reduced by 45 / 10 = 4.5 therefore , the overall average will be reduced by 4.5 the answer is c .
Answer: c
Problem: after 2 games , team b had an average of 61.5 points per game . if it got only 47 points in game 3 , how many more points does it need to score to get its total above 500 ?
Options: a ) 805 , b ) 740 , c ) 670 , d ) 530 , e ) 330
Rationale: ( 2 * 61.5 ) + 47 + x > 500 123 + 47 + x > 500 170 + x > 500 = > x > 330 option e
Answer: e
Problem: calculate the number of bricks , each measuring 25 cm * 15 cm * 8 cm required to construct a wall of dimensions 10 m * 4 m * 5 m when 10 % of its volume is occupied by mortar ?
Options: a ) 3388 , b ) 27678 , c ) 6000 , d ) 2977 , e ) 27681
Rationale: 10 * 4 / 100 * 5 * 90 / 100 = 25 / 100 * 15 / 100 * 8 / 100 * x 10 * 20 * 90 = 15 * 2 * x = > x = 6000 answer : c
Answer: c
Problem: a class has a ratio of 3 : 6 : 7 of children with red hair , blonde hair and black hair respectively . if the class has 9 kids with red hair , how many kids are in the class ?
Options: a ) 99 , b ) 48 , c ) 16 , d ) 22 , e ) 60
Rationale: since there are 9 children with red hair , and the class exists in a 3 : 6 : 7 ratio , there are 3 times more children than the unit ratio . therefore 9 red - haired children 3 * ( 6 ) blonde - haired children 3 * ( 7 ) black - haired children 9 + 18 + 21 = 48 alternatively , a part - to - whole ratio can be set up . since the whole is 3 + 6 + 7 = 16 , 9 / 3 = x / 16 3 * 16 = 48 answer : b )
Answer: b
Problem: a certain club has 10 members , including parry . one of the 10 members is chosen at random to be the president , one of the remaining 9 members is to be chosen at random to be the secretary , and one of the remaining 8 members is to be chosen at random to be the treasurer . what is the probability that parry will be either the member chose to be secretary or the member chose to be treasurer ?
Options: a ) 1 / 720 , b ) 1 / 80 , c ) 1 / 10 , d ) 1 / 9 , e ) 1 / 5
Rationale: in order to become a secretary the person can not be chosen as a president . this can be done in 9 / 10 ways . then to become a secretary this person ' s probability is 1 / 9 probability to become a secretary = ( 9 / 10 * 1 / 9 ) = 1 / 10 similar concept for treasurer . probability of not president = 9 / 10 , probability of not secretary = 8 / 9 , probability of treasurer 1 / 8 probability to become a treasurer = ( 9 / 10 * 8 / 9 * 1 / 8 ) = 1 / 10 since , the problem is saying parry secretary or treasurer it is addition : 1 / 10 + 1 / 10 = 1 / 5 = e
Answer: e
Problem: a millionaire bought a job lot of hats 1 / 4 of which were brown . the millionaire sold 2 / 3 of the hats including 4 / 5 of the brown hats . what fraction of the unsold hats were brown .
Options: a ) 1 / 60 , b ) 1 / 15 , c ) 3 / 20 , d ) 3 / 5 , e ) 3 / 4
Rationale: if initially he had x hats . x / 4 were brown . sold x * 2 / 3 hats . left with x / 3 hats . sold ( x / 4 ) * ( ( 4 / 5 ) = x / 5 brown hats and left with x / 4 - x / 5 = x / 20 brown hats . fraction of brown hats in total remaining hats = ( x / 20 ) / ( x / 3 ) = 3 / 20 hats answer : c
Answer: c
Problem: if three machines working at the same rate can do 3 / 4 of a job in 30 minutes , how many minutes would it take two machines working at the same rate to do 3 / 5 of the job ?
Options: a ) 36 , b ) 60 , c ) 75 , d ) 80 , e ) 100
Rationale: "using the std formula m 1 d 1 h 1 / w 1 = m 2 d 2 h 2 / w 2 substituting the values we have 3 * 1 / 2 * 4 / 3 = 2 * 5 / 3 * x ( converted 30 min into hours = 1 / 2 ) 2 = 10 / 3 * x x = 3 / 5 hour so 36 minutes answer : a"
Answer: a
Problem: { ( 481 + 426 ) 2 - 4 x 481 x 426 } = ?
Options: a ) 3025 , b ) 4200 , c ) 3060 , d ) 3210 , e ) none of these
Rationale: "explanation : here , the given statement is like ( a + b ) 2 - 4 ab where a = 481 and b = 426 ( a + b ) 2 - 4 ab = ( a 2 + 2 ab + b 2 ) - 4 ab = a 2 - 2 ab + b 2 = ( a - b ) 2 hence { ( 481 + 426 ) 2 - 4 x 481 x 426 } = ( 481 - 426 ) 2 = 552 = 3025 . answer : option a"
Answer: a
Problem: working together at their respective constant rates , machine a and machine b can produce 900 units in 8 hours . working alone , machine b would complete that same output in 50 % more time . if machine a were to work on its own for an 8 - hour shift , what percent of the 900 unit total would it produce ?
Options: a ) 30 , b ) 33 , c ) 50 , d ) 67 , e ) 75
Rationale: "1 / a + 1 / b = 1 / t 1 / a + 1 / 12 = 1 / 8 ( 50 % more of 8 is 12 ) 1 / a = 1 / 24 machine a can produce 900 units in 24 hrs , so it can produce 900 * 8 / 24 = 300 units is 8 hrs . 300 is 30 % of 900 . a is the answer"
Answer: a
Problem: sum of the squares of 3 no . ' s is 138 and the sum of their products taken two at a time is 131 . find the sum ?
Options: a ) 20 , b ) 24 , c ) 26 , d ) 28 , e ) 30
Rationale: ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 138 + 2 * 131 a + b + c = √ 400 = 20 a
Answer: a
Problem: a not - so - good clockmaker has four clocks on display in the window . clock # 1 loses 10 minutes every hour . clock # 2 gains 15 minutes every hour relative to clock # 1 ( i . e . , as clock # 1 moves from 12 : 00 to 1 : 00 , clock # 2 moves from 12 : 00 to 1 : 15 ) . clock # 3 loses 20 minutes every hour relative to clock # 2 . finally , clock # 4 gains 20 minutes every hour relative to clock # 3 . if the clockmaker resets all four clocks to the correct time at 12 noon , what time will clock # 4 display after 6 actual hours ( when it is actually 6 : 00 pm that same day ) ?
Options: a ) 5 : 00 , b ) 5 : 34 , c ) 5 : 42 , d ) 6 : 00 , e ) 6 : 24
Rationale: "c 1 loses 15 minutes every hour . so after 60 minutes have passed , c 1 displays that 60 - 15 = 45 minutes have passed . c 2 gains 15 minutes for every 60 minutes displayed on c 1 . thus , the time displayed on c 2 is 75 / 60 = 5 / 4 the time displayed on c 1 . so after 60 minutes have passed , c 2 displays the passing of ( 5 / 4 * 45 ) minutes . c 3 loses 20 minutes for every 60 minutes displayed on c 2 . thus , the time displayed on c 3 is 40 / 60 = 2 / 3 the time displayed on c 2 . so after 60 minutes have passed , c 3 displays the passing of ( 2 / 3 * 5 / 4 * 45 ) minutes . c 4 gains 20 minutes for every 60 minutes displayed on c 3 . thus , the time displayed on c 4 is 80 / 60 = 4 / 3 the time displayed on clock 3 . so after 60 minutes have passed , c 4 displays the passing of 4 / 3 * 2 / 3 * 5 / 4 * 45 = 50 minutes . c 4 loses 10 minutes every hour . in 6 hours , c 4 will lose 6 * 10 = 60 minutes = 1 hour . since the correct time after 6 hours will be 6 pm , c 4 will show a time of 6 - 1 = 6 pm . the correct answer is d ."
Answer: d
Problem: if x and y are positive integers and 11 + x + y + xy = 25 , what is the value of x + y ?
Options: a ) 3 , b ) 5 , c ) 6 , d ) 8 , e ) 9
Rationale: "try each answer choices . for a : 11 + 3 + xy = 25 ; xy = 11 ( impossible , 11 prime number . 1 + 11 does n ' t equal 3 ) for b : 11 + 5 + xy = 25 ; xy = 9 ( no combination of xy = 9 and x + y = 5 ) for c : 11 + 6 + xy = 25 ; xy = 8 ( x + y = 6 ; x = 2 , y = 4 or x = 4 , y = 2 ) for d : 11 + 8 + xy = 25 ; xy = 6 ( no combination of xy = 6 and x + y = 8 ) for e : 11 + 9 + xy = 25 ; xy = 5 ( impossible , 5 prime number . 1 + 5 does n ' t equal 9 ) therefore , answer c ."
Answer: c
Problem: a 6000 liter tank , half - full of water is being filled from a pipe with a flow rate of 1 kiloliter every 2 minutes . at the same time , the tank is losing water from two drains at a rate of 1 kiloliter every 4 minutes and every 6 minutes . how many minutes does it take to fill the tank completely ?
Options: a ) 8 , b ) 12 , c ) 18 , d ) 24 , e ) 36
Rationale: "in : we have : 1,000 / 2 min = 500 litres per minute out : we have : 1,000 / 4 + 1,000 / 6 then do : in - out to figure out the net inflow per minute ( you get 83.3 ) . then divide the total number of litres you need ( 3,000 by that net inflow to get the minutes ) - 36 min . answer e ."
Answer: e
Problem: rs . 1040 is divided amongst a , b , c so that 2 times a ' s share , 3 times b ' s share and 4 times c ' s share are all equal . find a ' s share ?
Options: a ) 177 , b ) 150 , c ) 817 , d ) 480 , e ) 616
Rationale: a + b + c = 1040 2 a = 3 b = 4 c = x a : b : c = 1 / 2 : 1 / 3 : 1 / 4 = 6 : 4 : 3 6 / 13 * 1040 = rs . 480 answer : d
Answer: d
Problem: salesperson a ' s compensation for any week is $ 300 plus 6 percent of the portion of a ' s total sales above $ 1,000 for that week . salesperson b ' s compensation for any week is 8 percent of b ' s total sales for that week . for what amount of total weekly sales would both salespeople earn the same compensation ?
Options: a ) $ 21,000 , b ) $ 18,000 , c ) $ 12,000 , d ) $ 8000 , e ) $ 4000
Rationale: "300 + 0.06 ( x - 1000 ) = 0.08 x 0.02 x = 240 x = $ 12,000 the answer is c ."
Answer: c
Problem: if the number is decreased by 5 and divided by 7 the result is 7 . what would be the result if 6 is subtracted and divided by 8 ?
Options: a ) 4 , b ) 6 , c ) 8 , d ) 5 , e ) 3
Rationale: explanation : let the number be x . then , ( x - 5 ) / 7 = 7 = > x - 5 = 49 x = 54 . : ( x - 6 ) / 8 = ( 54 - 6 ) / 8 = 6 answer : option b
Answer: b
Problem: ( x + 5 ) is a factor in x ^ 2 - mx - 40 . what is the value of m ?
Options: a ) 1 , b ) 2 , c ) 4 , d ) 3 , e ) 5
Rationale: "i solved the second degree equation and found it like this : x ^ 2 - mx - 40 = 0 ( x - 8 ) ( x + 5 ) = 0 x = 8 or x = - 5 substituting both values for x in the equation we find : x ^ 2 - mx - 40 = > ( - 5 ) ^ 2 - m ( - 5 ) = 40 = > 25 + 5 m = 40 = > 5 m = 40 - 25 = 15 = > m = 3 and with 8 , using a similar process we end up with : ( 8 ) ^ 2 - m ( 8 ) = 40 - 8 m = 40 - 64 = - 24 m = 3 ao , ans d"
Answer: d
Problem: in a rectangular coordinate system , what is the area of a quadrilateral whose vertices have the coordinates ( 4 , - 3 ) , ( 4 , 7 ) , ( 10 , 4 ) , ( 10 , - 7 ) ?
Options: a ) 51 , b ) 54 , c ) 57 , d ) 60 , e ) 63
Rationale: "by graphing the points , we can see that this figure is a trapezoid . a trapezoid is any quadrilateral that has one set of parallel sides , and the formula for the area of a trapezoid is : area = ( 1 / 2 ) × ( base 1 + base 2 ) × ( height ) , where the bases are the parallel sides . we can now determine the area of the quadrilateral : area = 1 / 2 × ( 10 + 11 ) × 6 = 63 . the answer is e ."
Answer: e
Problem: a sum is divided among b , c and d in such a way that for each rupee b gets , c gets 150 paisa and d gets 50 paisa . if the share of c is rs . 40 , what is the total amount ?
Options: a ) 70 , b ) 75 , c ) 80 , d ) 85 , e ) 90
Rationale: b : c : d = 100 : 150 : 100 20 : 30 : 10 30 - - - 40 60 - - - ? = > 80 answer : c
Answer: c
Problem: find the average marks of all the students in 2 separate classes , if the average marks of students in the first class of 58 students is 67 and that of another class of 52 students is 82 .
Options: a ) 74.1 , b ) 56.3 , c ) 67.8 , d ) 79.7 , e ) 19.4
Rationale: "sum of the marks for the class of 58 students = 58 * 67 = 3886 sum of the marks for the class of 52 students = 52 * 82 = 4264 sum of the marks for the class of 110 students = 3886 + 4262 = 8150 average marks of all the students = 4200 / 80 = 74.1 answer : a"
Answer: a