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a ) 2 % , b ) 1 % , c ) 6 % , d ) 5 % , e ) 8 %
d
multiply(divide(divide(const_3, 3), add(multiply(const_3, 3), add(const_0_25, const_0_25))), const_100)
a sum of 14,400 amounts to 16,600 in 3 years at the rate of simple interest . what is the rate of interest ?
"d 5 % s . i . = ( 16600 - 14400 ) = 2200 . rate = ( 100 x 2200 ) / ( 14400 x 3 ) % = 5 %"
a = 3 / 3 b = 3 * 3 c = const_0_25 + const_0_25 d = b + c e = a / d f = e * 100
a ) 20 , b ) 30 , c ) 35 , d ) 40 , e ) 55
b
divide(subtract(75, 66), subtract(const_1, divide(70, const_100)))
a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 70 percent of books that were loaned out are returned and there are 66 books in the special collection at that time , how many books of the special collection were loaned out during that month ?
total = 75 books . 65 % of books that were loaned out are returned - - > 100 % - 70 % = 30 % of books that were loaned out are not returned . now , there are 66 books , thus 76 - 68 = 7 books are not returned . { loaned out } * 0.30 = 7 - - > { loaned out } = 30 . answer : b .
a = 75 - 66 b = 70 / 100 c = 1 - b d = a / c
a ) 7 meters , b ) 9 meters , c ) 8 meters , d ) 5 meters , e ) 3 meters
a
multiply(divide(168, multiply(multiply(multiply(const_2, const_3), const_10), 400)), const_1000)
if a tire rotates at 400 revolutions per minute when the car is traveling 168 km / h , what is the circumference of the tire ?
"400 rev / minute = 400 * 60 rev / 60 minutes = 24,000 rev / hour 24,000 * c = 168,000 m : c is the circumference c = 3 meters correct answer a"
a = 2 * 3 b = a * 10 c = b * 400 d = 168 / c e = d * 1000
a ) 52 , b ) 49 , c ) 95 , d ) 88 , e ) 66
d
add(multiply(add(15, subtract(15, 3)), 3), 7)
in a bus left side are 15 seats available , 3 few seats in right side because in rear exit door . each seat hold 3 people . in addition , there is a seat back can sit 7 people all together . how many people can sit in a bus ?
"right side = 15 seat left side = 15 - 3 ( 3 few seat in right side ) = 12 seat total = 15 + 12 = 27 people can seat in 27 seat = 27 * 3 = 81 people can seat in last seat = 7 total people can seat = 81 + 7 = 88 answer : d"
a = 15 - 3 b = 15 + a c = b * 3 d = c + 7
a ) 1 / 7 , b ) 1 / 6 , c ) 1 / 5 , d ) 1 / 4 , e ) 1 / 3
a
inverse(add(divide(const_12, 2), const_1))
a worker ' s take - home pay last year was the same each month , and she saved the same fraction of her take - home pay each month . the total amount of money that she had saved at the end of the year was 2 times the amount of that portion of her monthly take - home pay that she did not save . if all the money that she saved last year was from her take - home pay , what fraction of her take - home pay did she save each month ?
"let x be the fraction of her take - home pay that the worker saved . let p be the monthly pay . 12 xp = 2 ( 1 - x ) p 12 xp = 2 p - 2 xp 14 xp = 2 p x = 1 / 7 the answer is a ."
a = 12 / 2 b = a + 1 c = 1/(b)
a ) 12 , b ) 18 , c ) 6 , d ) 1 , e ) 2
a
multiply(divide(multiply(multiply(75, 20), 50), multiply(multiply(100, 25), 30)), 12)
if a certain number of labours can dig earth 100 m deep , 25 m long , and 30 m broad in 12 days , then find the number of days the same number of labours will require to dig earth 75 m deep , 20 m long and 50 m broad .
explanation : more number of days means more length , more depth and more width . hence , its a direct proportion . ( 100 * 25 * 30 ) : ( 75 * 20 * 50 ) : : 12 : x 75000 : 75000 : : 12 : x x = 12 answer : a
a = 75 * 20 b = a * 50 c = 100 * 25 d = c * 30 e = b / d f = e * 12
a ) 28.5 , b ) 27.675 , c ) 30 , d ) 34 % , e ) none of these
d
subtract(divide(multiply(add(const_100, 27.3), const_100), subtract(const_100, 5)), const_100)
a shopkeeper sold an article offering a discount of 5 % and earned a profit of 27.3 % . what would have been the percentage of profit earned if no discount had been offered ?
"giving no discount to customer implies selling the product on printed price . suppose the cost price of the article is 100 . then printed price = 100 Γ£ β€” ( 100 + 27.3 ) / ( 100 Γ’ Λ† ’ 5 ) = 100 Γ£ β€” 127.3 / 95 = 134 hence , required % profit = 134 Γ’ € β€œ 100 = 34 % answer d"
a = 100 + 27 b = a * 100 c = 100 - 5 d = b / c e = d - 100
a ) 16 % , b ) 45 % , c ) 32 % , d ) 40 % , e ) 28 %
e
multiply(divide(subtract(64, 45), 64), const_100)
in town x , 64 percent of the population are employed , and 45 percent of the population are employed males . what percent of the employed people in town x are females ?
"answer b male - employed - 64 % female - employed - 19 % total employed 64 % means total unemployed = 36 % therefore d and e are not correct because female unemployed has to be less than 36 % female - unemployed = 32 % male unemployed = 4 % 48 % + 4 % = 52 % 16 % + 32 % = 48 % plug in a and b in place of 32 % and the sum is not 100 % 19 / 64 = 28 e"
a = 64 - 45 b = a / 64 c = b * 100
a ) 33488 , b ) 63809 , c ) 60000 , d ) 37887 , e ) 77824
c
divide(multiply(multiply(add(const_1, const_4), const_1000), 2), 3)
x and y invested in a business . they earned some profit which they divided in the ratio of 2 : 3 . if x invested rs . 40,000 , the amount invested by y is
"explanation : suppose y invested rs . y . then 40000 / y = 2 / 3 or y = 60000 . answer : c ) 60000"
a = 1 + 4 b = a * 1000 c = b * 2 d = c / 3
a ) 100.6 kg , b ) 142 kg , c ) 135 kg , d ) 110.8 kg , e ) 114 kg
b
add(multiply(divide(135, add(3, 5)), 5), multiply(divide(145, add(2, 3)), 2))
135 kg of an alloy a is mixed with 145 kg of alloy b . if alloy a has lead and tin in the ratio 3 : 5 and alloy b has tin and copper in the ratio 2 : 3 , then the amount of tin in the new alloy is ?
"quantity of tin in 135 kg of a = 135 * 5 / 8 = 84 kg quantity of tin in 145 kg of b = 145 * 2 / 5 = 58 kg quantity of tin in the new alloy = 84 + 58 = 142 kg answer is b"
a = 3 + 5 b = 135 / a c = b * 5 d = 2 + 3 e = 145 / d f = e * 2 g = c + f
a ) $ 340 , b ) $ 360 , c ) $ 380 , d ) $ 400 , e ) $ 420
d
subtract(520, multiply(subtract(820, 520), divide(2, 5)))
a sum of money lent out at s . i . amounts to a total of $ 520 after 2 years and to $ 820 after a further period of 5 years . what was the initial sum of money that was invested ?
"s . i for 5 years = $ 820 - $ 520 = $ 300 the s . i . is $ 60 / year s . i . for 2 years = $ 120 principal = $ 520 - $ 120 = $ 400 the answer is d ."
a = 820 - 520 b = 2 / 5 c = a * b d = 520 - c
a ) 167.5 , b ) 150 , c ) 225 , d ) 112.5 , e ) 212.5
d
divide(subtract(divide(multiply(multiply(5000, 61), 2), const_100), divide(multiply(multiply(5000, 4), 2), const_100)), 2)
a person borrows rs . 5000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 61 ⁄ 4 % p . a for 2 years . find his gain in the transaction per year .
"explanation : the person borrows rs . 5000 for 2 years at 4 % p . a . simple interest simple interest that he needs to pay = prt / 100 = 5000 Γ— 4 Γ— 2 / 100 = 400 he also lends it at 6 1 ⁄ 4 % p . a for 2 years simple interest that he gets = prt / 100 = 5000 Γ— 254 Γ— 21 / 00 = 625 his overall gain in 2 years = rs . 625 - rs . 400 = rs . 225 his overall gain in 1 year = 2252 = rs . 112.5 answer : option d"
a = 5000 * 61 b = a * 2 c = b / 100 d = 5000 * 4 e = d * 2 f = e / 100 g = c - f h = g / 2
a ) 1 / 6 , b ) 4 / 9 , c ) 1 / 4 , d ) 5 / 9 , e ) 1 / 3
b
inverse(add(inverse(4), 2))
a small water pump would take 4 hours to fill an empty tank . a larger pump would take 1 / 2 hour to fill the same tank . how many hours would it take both pumps , working at their respective constant rates , to fill the empty tank if they began pumping at the same time ?
"rate of the small pump is 1 / 4 tank / hour rate of the larger pump is 1 / ( 1 / 2 ) or 2 tank / hour ; combined rate of the two pumps is 1 / 4 + 2 = 9 / 4 tank / hour , together they will fill the empty tank in 1 / ( 9 / 4 ) ) = 4 / 9 hours ( time = job / rate ) . answer : b"
a = 1/(4) b = a + 2 c = 1/(b)
a ) 5 minutes , b ) 6 minutes , c ) 8 minutes , d ) 9 / 2 minutes , e ) 10 minutes
d
divide(subtract(3, divide(3, divide(add(9, 3), subtract(3, 9)))), const_1)
a man cycling along the road noticed that every 3 minutes a bus overtakes him and every 9 minutes he meets an oncoming bus . if all buses and the cyclist move at a constant speed , what is the time interval between consecutive buses ?
"let ' s say the distance between the buses is d . we want to determine interval = \ frac { d } { b } , where b is the speed of bus . let the speed of cyclist be c . every 9 minutes a bus overtakes cyclist : \ frac { d } { b - c } = 9 , d = 9 b - 9 c ; every 3 minutes cyclist meets an oncoming bus : \ frac { d } { b + c } = 3 , d = 3 b + 3 c ; d = 9 b - 9 c = 3 b + 3 c , - - > b = 2 c , - - > d = 9 b - 9 b / 2 = 9 b / 2 . interval = \ frac { d } { b } = \ frac { 9 / 2 b } { b } = 9 / 2 answer : d ( 9 / 2 minutes ) ."
a = 9 + 3 b = 3 - 9 c = a / b d = 3 / c e = 3 - d f = e / 1
a ) 6 hours , b ) 8 hours , c ) 9 hours , d ) 10 hours , e ) 12 hours
b
inverse(subtract(divide(2, 2), inverse(divide(add(multiply(2, 3), 2), 3))))
a pump can fill a tank with water in 2 hours . because of a leak , it took 2 2 / 3 hours to fill the tank . the leak can drain all the water of the tank in
"explanation : solution 1 the pump can fill the tank in 2 hour . but , because of the leak , it took 2 2 / 3 hours to fill the tank . i . e . , the additional water filled by the pump in 2 / 3 hour is emptied by the leak in 2 2 / 3 = 8 / 3 hours . therefore , the water filled by pump in 2 bour can be emptied by the leak in 8 / 3 x 3 = 8 hours . i . e . , the leak can drain all the water of the tank in 8 hours . answer is b"
a = 2 / 2 b = 2 * 3 c = b + 2 d = c / 3 e = 1/(d) f = a - e g = 1/(f)
a ) 230 m , b ) 240 m , c ) 260 m , d ) 270 m , e ) none of the above
d
multiply(subtract(26, divide(250, multiply(const_0_2778, 72))), multiply(const_0_2778, 72))
a goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds . what is the length of the goods train ?
"speed = ( 72 x ( 5 / 18 ) ) m / sec = 20 m / sec . time = 26 sec . let the length of the train be x metres . then , ( ( x + 250 ) / 26 ) = 20 x + 250 = 520 x = 270 . d"
a = const_0_2778 * 72 b = 250 / a c = 26 - b d = const_0_2778 * 72 e = c * d
a ) 15 , b ) 60 , c ) 75 , d ) 90 , e ) 120
e
multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60)
if teena is driving at 55 miles per hour and is currently 7.5 miles behind joe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 22.5 miles ahead of joe ?
"this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . joe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 22.5 miles . teena can cover 7.5 miles in 30 mins teena will cover 22.5 miles in 90 mins so answer 30 + 90 = 120 mins . ( answer e )"
a = 55 - 40 b = a + 7 c = 55 - 40 d = b / c e = d * const_60
a ) a ) 70 , b ) b ) 77 , c ) c ) 55 , d ) d ) 81 , e ) e ) 82
c
multiply(divide(divide(multiply(21, add(21, const_1)), const_2), 21), 5)
what is the average of first 21 multiples of 5 ?
required average = 7 ( 1 + 2 + . . . . + 21 ) / 21 ( 5 / 21 ) x ( ( 21 x 22 ) / 2 ) ( because sum of first 21 natural numbers ) = 55 c
a = 21 + 1 b = 21 * a c = b / 2 d = c / 21 e = d * 5
a ) 321 , b ) 314 , c ) 316 , d ) 265 , e ) 162
b
subtract(subtract(560, divide(multiply(560, 20), const_100)), divide(multiply(subtract(560, divide(multiply(560, 20), const_100)), 30), const_100))
the sale price sarees listed for rs . 560 after successive discount is 20 % and 30 % is ?
"explanation : 560 * ( 80 / 100 ) * ( 70 / 100 ) = 314 answer : b"
a = 560 * 20 b = a / 100 c = 560 - b d = 560 * 20 e = d / 100 f = 560 - e g = f * 30 h = g / 100 i = c - h
a ) 1 / 200 , b ) 1 / 100 , c ) 90 / 2900 , d ) 17 / 2700 , e ) 1 / 20
d
divide(subtract(multiply(choose(45, const_1), choose(60, const_1)), choose(17, const_1)), multiply(choose(45, const_1), choose(60, const_1)))
45 girls and 60 boys are present at a party . there are 17 girl - boy are in relationship among them . if a girl and a boy is selected at random , what is the probability that they will be a couple ?
"in how many ways we can select a girl and a boy from 45 lady and 60 gentlemen ? in 45 * 60 = 2,700 ways . we have a total of 17 couples so , the probability of selecting a couple is 17 / 2,700 = 17 / 2700 . ans - d"
a = math.comb(45, 1) b = math.comb(60, 1) c = a * b d = math.comb(17, 1) e = c - d f = math.comb(45, 1) g = math.comb(60, 1) h = f * g i = e / h
a ) 2 days , b ) 4 1 / 2 days , c ) 6 1 / 2 days , d ) 7 3 / 4 days , e ) 10 days
c
add(multiply(12, 2), divide(12, 2))
a can complete a work in 12 days and b can do the same work in 8 days . if a after doing 2 days , leaves the work , find in how many days b will do the remaining work ?
the required answer = ( 12 - 2 ) * 8 / 12 = 80 / 12 = 6 1 / 2 days answer is c
a = 12 * 2 b = 12 / 2 c = a + b
a ) 2898 , b ) 1800 , c ) 500 , d ) 297 , e ) 435
b
divide(800, multiply(subtract(78, 2), const_0_2778))
a train 800 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 2 min , then the length of the tunnel is ?
"speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 2 min = 120 sec . let the length of the train be x meters . then , ( 800 + x ) / 120 = 65 / 3 x = 1800 m . answer : b"
a = 78 - 2 b = a * const_0_2778 c = 800 / b
a ) 5 , b ) 46 , c ) 52 , d ) 12 , e ) 3
e
divide(75, 25)
given that e and f are events such that p ( e ) = 25 , p ( f ) = 75 and p ( e n f ) = 75 , find p ( e | f ) and p ( f | e )
here , e and f are events p ( e | f ) = p ( enf ) / p ( f ) = 75 / 25 = 3 p ( f | e ) = p ( enf ) / p ( e ) = 75 / 75 = 1 . e )
a = 75 / 25
a ) 90 , b ) 60 , c ) 120 , d ) 180 , e ) 360
a
divide(divide(2500, const_1000), divide(100, const_3600))
a train 2500 m long can cross an electric pole in 100 sec and then find the speed of the train ?
"length = speed * time speed = l / t s = 2500 / 100 s = 25 m / sec speed = 25 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 90 kmph answer : a"
a = 2500 / 1000 b = 100 / 3600 c = a / b
a ) 5 , b ) 7 , c ) 11 , d ) 13 , e ) 11 * 455
e
divide(add(divide(subtract(multiply(floor(divide(630, 35)), 35), multiply(floor(divide(280, 35)), 35)), 35), const_1), const_2)
if integer k is equal to the sum of all even multiples of 35 between 280 and 630 , what is the greatest prime factor of k ?
"if we break down what the stem is asking what is the sum of all mult of 30 between 300 and 600 . using arithmetic progression to find n : 630 = 280 + ( n - 1 ) 35 350 + 35 = 35 n 385 = 35 n = > n = 11 the sum would be : 11 * mean mean = [ 630 + 280 ] / 2 = 455 11 * 455 = e"
a = 630 / 35 b = math.floor(a) c = b * 35 d = 280 / 35 e = math.floor(d) f = e * 35 g = c - f h = g / 35 i = h + 1 j = i / 2
a ) 50 , b ) 60 , c ) 40 , d ) 18 , e ) 22
b
multiply(divide(4, add(add(add(5, 7), 4), 8)), 360)
if 360 is divided into 4 parts proportional to 5 , 7 , 4 , 8 then the smallest part is :
given ratio = 5 : 7 : 4 : 8 , sum of ratio = 24 : . the smallest part = ( 360 * 4 / 24 ) = 60 answer : b
a = 5 + 7 b = a + 4 c = b + 8 d = 4 / c e = d * 360
a ) - 37680 , b ) 6530 , c ) 6630 , d ) 6730 , e ) 6830
a
subtract(negate(180), multiply(subtract(1320, 1070), divide(subtract(1320, 1070), subtract(1370, 1320))))
1370 , 1320 , 1070 , - 180 , - 6430 , x
"1370 - 50 * ( 5 ^ 0 ) = 1320 1320 - 50 * ( 5 ^ 1 ) = 1070 1070 - 50 * ( 5 ^ 2 ) = - 180 - 180 - 50 * ( 5 ^ 3 ) = - 6430 - 6430 - 50 * ( 5 ^ 4 ) = - 37680 answer : a ."
a = negate - (
a ) 25 inches , b ) 77 inches , c ) 66 inches , d ) 97 inches , e ) 66 inches
a
divide(add(multiply(10, const_12), 5), 5)
a scale 10 ft . 5 inches long is divided into 5 equal parts . find the length of each part .
"explanation : total length of scale in inches = ( 10 * 12 ) + 5 = 125 inches length of each of the 5 parts = 125 / 5 = 25 inches answer : a"
a = 10 * 12 b = a + 5 c = b / 5
a ) 54 kmph , b ) 60 kmph , c ) 66 kmph , d ) 72 kmph , e ) 84 kmph
e
divide(divide(subtract(150, multiply(multiply(6, const_0_2778), 6)), 6), const_0_2778)
a train 150 m long passes a man , running at 6 kmph in the direction opposite to that of the train , in 6 seconds . the speed of the train is
"speed of train relative to man : 150 / 6 * 18 / 5 km / hr = 90 km / hr let speed of train = x therefore x + 6 = 90 x = 90 - 6 x = 84 km / hr answer : e"
a = 6 * const_0_2778 b = a * 6 c = 150 - b d = c / 6 e = d / const_0_2778
a ) 4 hrs , b ) 4.5 hrs , c ) 5.4 hrs , d ) 6 hrs . , e ) none of these
a
divide(add(60, multiply(20, 1)), subtract(60, 40))
a thief runs away from a with a speed of 20 km / hr . a police from a station 60 km away started chasing him after 1 hr . with a speed of 40 km / hr . time taken by the police to catch the thief is :
explanation : solution : distance traveled by the thief in 1 hr . = 20 km . total distance of them after 1 hr = ( 60 + 20 ) km = 80 km . their relative speed = ( 40 - 20 ) = 20 km / hr . time taken to catch the thief = 80 / 20 = 4 hrs . answer : a
a = 20 * 1 b = 60 + a c = 60 - 40 d = b / c
a ) 9 , b ) 2 , c ) 122 , d ) 6 , e ) 1
b
divide(sqrt(16), 2)
what is the square root of 16 , divided by 2 ?
"square root is a number times itself square root of 16 = 4 , 4 / 2 = 2 ( b ) 2"
a = math.sqrt(16) b = a / 2
a ) 0.47 , b ) 47 , c ) 470 , d ) 4700 , e ) none of these
c
multiply(4.7, add(add(13.26, 9.43), 77.31))
the value of ( 4.7 Γ— 13.26 + 4.7 Γ— 9.43 + 4.7 Γ— 77.31 ) is :
solution given expression = 4.7 Γ— ( 13.26 + 9.43 + 77.31 ) = 4.7 Γ— 100 = 470 . answer c
a = 13 + 26 b = a + 77 c = 4 * 7
a ) 1.0 , b ) 0.5 , c ) 0.75 , d ) 0.425 , e ) 0.3
d
multiply(divide(divide(500, const_1000), divide(60, const_60)), subtract(const_1, divide(15, const_100)))
an ant walks an average of 500 meters in 60 minutes . a beetle walks 15 % less distance at the same time on the average . assuming the beetle walks at her regular rate , what is its speed in km / h ?
"the ant walks an average of 500 meters in 60 minutes 500 meters in 1 hours the beetle walks 15 % less distance = 500 - 75 = 425 meters in 60 minutes 0.425 km in 1 hours speed = 0.425 * 1 = 0.425 km / h i guess option d should be 0.425"
a = 500 / 1000 b = 60 / const_60 c = a / b d = 15 / 100 e = 1 - d f = c * e
a ) 37.5 , b ) 75 , c ) 100 , d ) 150 , e ) 175
d
divide(subtract(multiply(20, const_2), 10), subtract(subtract(subtract(1, divide(20, const_100)), multiply(subtract(1, divide(20, const_100)), divide(1, const_4))), multiply(const_2, divide(20, const_100))))
a tank holds x gallons of a saltwater solution that is 20 % salt by volume . one fourth of the water is evaporated , leaving all of the salt . when 10 gallons of water and 20 gallons of salt are added , the resulting mixture is 33 1 / 3 % salt by volume . what is the value of x ?
"original mixture has 20 % salt and 80 % water . total = x out of which salt = 0.2 x and water = 0.8 x now , 1 / 4 water evaporates and all salt remains . so what remains is 0.2 x salt and 0.6 x water . now 20 gallons salt is added and 10 gallons of water is added . so salt now becomes - > ( 0.2 x + 20 ) and water - - > ( 0.6 x + 10 ) amount of salt is 33.33 % of total . so amount of water is 66.66 % . so salt is half of the volume of water . so ( 0.2 x + 20 ) = ( 0.6 x + 10 ) / 2 , solving , x = 150 . answer : d"
a = 20 * 2 b = a - 10 c = 20 / 100 d = 1 - c e = 20 / 100 f = 1 - e g = 1 / 4 h = f * g i = d - h j = 20 / 100 k = 2 * j l = i - k m = b / l
a ) 156 , b ) 162 , c ) 325 , d ) 324 , e ) 326
c
add(multiply(floor(divide(680, 50)), multiply(const_12, const_2)), floor(divide(subtract(680, multiply(floor(divide(680, 50)), 50)), 2.30)))
roses can be purchased individually for $ 2.30 , one dozen for $ 36 , or two dozen for $ 50 . what is the greatest number of roses that can be purchased for $ 680 ?
"buy as many $ 50 deals as possible . we can by 650 / 50 = 13 two dozen roses , thus total of 13 * 24 = 312 roses . we are left with 680 - 650 = $ 30 . we can buy 30 / 2.3 = ~ 13 roses for that amount . total = 312 + 13 = 325 . answer : c ."
a = 680 / 50 b = math.floor(a) c = 12 * 2 d = b * c e = 680 / 50 f = math.floor(e) g = f * 50 h = 680 - g i = h / 2 j = math.floor(i) k = d + j
a ) 65 , b ) 66 , c ) 67 , d ) 131 , e ) 130
e
add(add(const_1, 65), 65)
in the land of oz only one or two - letter words are used . the local language has 65 different letters . the parliament decided to forbid the use of the seventh letter . how many words have the people of oz lost because of the prohibition ?
the answer to the question is indeed e . the problem with above solutions is that they do not consider words like aa , bb , . . . the number of 1 letter words ( x ) that can be made from 65 letters is 65 ; the number of 2 letter words ( xx ) that can be made from 65 letters is 65 * 65 , since each x can take 65 values . total : 65 + 65 * 65 . similarly : the number of 1 letter words ( x ) that can be made from 64 letters is 64 ; the number of 2 letter words ( xx ) that can be made from 64 letters is 64 * 64 , since each x can take 64 values . total : 64 + 64 * 64 . the difference is ( 65 + 65 * 65 ) - ( 64 + 64 * 64 ) = 130 . answer : e .
a = 1 + 65 b = a + 65
a ) 330 , b ) 300 , c ) 252 , d ) 225 , e ) 210
e
multiply(multiply(add(const_4, const_2), 7), 5)
baker ram is storing cake boxes in cartons that measure 25 inches by 42 inches by 60 inches . if the measurement of each cake box is 8 inches by 7 inches by 5 inches , then what is the maximum number of cake boxes that can be placed in each carton ?
baker ram is storing cake boxes in cartons the dimensions of the carton are 25 x 42 x 60 and those of each cake box are 8 x 7 x 5 . to maximize number of boxes try and fit maximum into the base of the box . if you take the base as 25 x 42 and fit 5 x 6 boxes of 5 x 7 , then the total number of boxes is 5 x 6 x integer ( 60 / 8 ) = 5 x 6 x 7 = 210 . alternatively you could try taking the base as 60 x 42 and fit 12 x 6 boxes of 5 x 7 , then the total number of boxes is 12 x 6 x integer ( 25 / 8 ) = 12 x 6 x 3 = 216 . but since that is not an answer option , i go with ( e ) 210 .
a = 4 + 2 b = a * 7 c = b * 5
a ) 7 % , b ) 10 % , c ) 13 % , d ) 16 % , e ) 19 %
c
multiply(divide(add(multiply(40, divide(5, const_100)), 4.5), add(40, add(4.5, 5.5))), const_100)
a 40 - liter solution of alcohol and water is 5 percent alcohol . if 4.5 liters of alcohol and 5.5 liters of water are added to this solution , what percent of the solution produced is alcohol ?
the percent of alcohol in the solution is ( 0.05 ( 40 ) + 4.5 ) / 50 = 6.5 / 50 = 13 % the answer is c .
a = 5 / 100 b = 40 * a c = b + 4 d = 4 + 5 e = 40 + d f = c / e g = f * 100
a ) 12 , b ) 16 , c ) 17 , d ) 20 , e ) 21
c
subtract(20, 4)
a snail , climbing a 20 feet high wall , climbs up 4 feet on the first day but slides down 2 feet on the second . it climbs 4 feet on the third day and slides down again 2 feet on the fourth day . if this pattern continues , how many days will it take the snail to reach the top of the wall ?
"total transaction in two days = 4 - 2 = 2 feet in 16 days it will climb 16 feet on the 17 th day , the snail will climb 4 feet , thus reaching the top therefore , total no of days required = 17 answer : c"
a = 20 - 4
['a ) 2', 'b ) 3', 'c ) 6', 'd ) 12', 'e ) 24']
b
divide(divide(divide(divide(divide(divide(432, const_2), const_2), const_2), const_2), const_3), const_3)
if n is the smallest integer such that 432 times n is the square of an integer , what is the value of n ?
the prime factorization of a square has to have even powers of all its prime factors . if the original number has a factor , say of 7 , then when it ’ s squared , the square will have a factor of 7 ^ 2 . another way to say that is : any positive integer all of whose prime factors have even powers must be a perfect square of some other integer . look at the prime factorization of 432 432 = ( 2 ^ 4 ) * ( 3 ^ 3 ) the factor of 2 already has an even power β€” - that ’ s all set . the factor of 3 currently has an odd power . if n = 3 , then 432 * n would have an even power of 2 and an even power of 3 ; therefore , it would be a perfect square . thus , n = 3 is a choice that makes 432 * n a perfect square . answer : b .
a = 432 / 2 b = a / 2 c = b / 2 d = c / 2 e = d / 3 f = e / 3
a ) 72 , b ) 23 , c ) 29 , d ) 26 , e ) 13
a
multiply(36, const_2)
the average of the marks of 12 students in a class is 36 . if the marks of each student are doubled , find the new average ?
"sum of the marks for the 12 students = 12 * 36 = 432 . the marks of each student are doubled , the sum also will be doubled . the new sum = 432 * 2 = 864 . so , the new average = 864 / 12 = 72 . answer : a"
a = 36 * 2
a ) 21 , b ) 28 , c ) 90 , d ) 85 , e ) 11
d
subtract(divide(8925, 85), 20)
a trader sells 85 meters of cloth for rs . 8925 at the profit of rs . 20 per metre of cloth . what is the cost price of one metre of cloth ?
"explanation : sp of 1 m of cloth = 8925 / 85 = rs . 105 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 105 - rs . 20 = rs . 85 . answer : d"
a = 8925 / 85 b = a - 20
a ) 0.6 % , b ) 0.5 % , c ) 0.1 % , d ) 0.2 % , e ) 0.7 %
e
multiply(divide(divide(20, 2), 1500), const_100)
if the simple interest on $ 1500 increase by $ 20 , when the time increase by 2 years . find the rate percent per annum .
applying the above formula , we have 20 = 1500 * r * 2 / 100 r = 20 * 100 / 1500 * 2 = 0.7 % . answer is e
a = 20 / 2 b = a / 1500 c = b * 100
a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70
b
divide(multiply(8.40, const_100), subtract(add(25, const_100), divide(multiply(add(30, const_100), subtract(const_100, 20)), const_100)))
a man sells an article at a profit of 25 % . if he had bought it at 20 % less and sold it for rs . 8.40 less , he would have gained 30 % . find the cost of the article .
"let c . p = 100 gain = 25 % s . p = 125 supposed c . p = 80 gain = 30 % s . p = ( 130 * 80 ) / 100 = 104 diff = ( 125 - 104 ) = 21 diff 21 when c . p = 100 then diff 8.40 when c . p = ( 100 * 8.40 ) / 21 = 40 answer : b"
a = 8 * 40 b = 25 + 100 c = 30 + 100 d = 100 - 20 e = c * d f = e / 100 g = b - f h = a / g
a ) 13 , b ) 5 , c ) 8 , d ) 11 , e ) 12
b
divide(subtract(145, power(29, const_2)), const_2)
the sum of two numbers is 29 and the difference of their squares is 145 . the difference between the number is
"explanation : let the numbers be x and y . we know that , ( x βˆ’ y ) = x 2 βˆ’ y 2 / ( x + y ) = 145 / 29 = 5 correct option : b"
a = 29 ** 2 b = 145 - a c = b / 2
a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20
b
subtract(divide(50, 2), add(const_4, const_4))
the cube root of what integer power of 2 is closest to 50 ?
integer powers of 2 = 2 , 2 ^ 2 , 2 ^ 3 , 2 ^ 4 , 2 ^ 5 , 2 ^ 6 , 2 ^ 7 etc . let , cube root of x is closest to 50 i . e . x is closest to cube of 50 = 50 * 50 * 50 = 125000 but x must be equivalent to an integer power of 2 2 ^ 10 = 1024 2 ^ 7 = 128 i . e . 2 ^ 17 = 128 * 1024 = 128000 ( approx . ) i . e . 2 ^ 18 = 256 * 1024 = 256000 ( approx . ) i . e . 2 ^ 16 = 64 * 1024 = 64000 ( approx . ) i . e . 2 ^ 17 which is closest to 125000 i . e . integer power of 2 must be 7 answer : option b
a = 50 / 2 b = 4 + 4 c = a - b
a ) s . 1000 , b ) s . 1009 , c ) s . 1007 , d ) s . 1357.1 , e ) s . 1002
d
divide(multiply(190, const_100), subtract(add(const_100, 4), subtract(const_100, 10)))
a watch was sold at a loss of 10 % . if it was sold for rs . 190 more , there would have been a gain of 4 % . what is the cost price ?
"explanation : 90 % 104 % - - - - - - - - 14 % - - - - 190 100 % - - - - ? = > rs . 1357.1 answer : d"
a = 190 * 100 b = 100 + 4 c = 100 - 10 d = b - c e = a / d
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12
b
subtract(9, divide(log(divide(power(3, 9), 19683)), log(multiply(3, 4))))
if 3 ^ x * 4 ^ y = 19683 and x – y = 9 , then x = ?
since 19,683 is an odd number , it is not a multiple of 4 . this means that y must equal zero . since x - y = 9 and y = 0 , then x = 9 . the answer is b .
a = 3 ** 9 b = a / 19683 c = math.log(b) d = 3 * 4 e = math.log(d) f = c / e g = 9 - f
a ) 1.6 cm , b ) 2.4 cm , c ) 4.8 cm , d ) 2.16 cm , e ) 3.2 cm
a
divide(multiply(2, 8), const_10)
in triangle pqr , the angle q = 90 degree , pq = 2 cm , qr = 8 cm . x is a variable point on pq . the line through x parallel to qr , intersects pr at y and the line through y , parallel to pq , intersects qr at z . find the least possible length of xz
"look at the diagram below : now , in case when qy is perpendicular to pr , two right triangles pqr and pqy are similar : qy : qp = qr : pr - - > qy : 2 = 8 : 10 - - > qy = 1.6 . answer : a ."
a = 2 * 8 b = a / 10
a ) 6 rs , b ) 2 rs , c ) 4 rs , d ) 9 rs , e ) 14 rs
e
divide(210, multiply(const_3, 5))
5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 210 only . men Γ’ € β„’ s wages are ?
"5 m = xw = 8 b 5 m + xw + 8 b - - - - - 210 rs . 5 m + 5 m + 5 m - - - - - 210 rs . 15 m - - - - - - 210 rs . = > 1 m = 14 rs . answer : e"
a = 3 * 5 b = 210 / a
a ) 62 % , b ) 68 % , c ) 72 % , d ) 76 % , e ) 80 %
a
multiply(divide(add(const_100, 10), add(add(const_100, 10), const_100)), const_100)
in may mrs lee ' s earnings were 60 percent of the lee family ' s total income . in june mrs lee earned 10 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 = 60 in may rest of the family ' s income = 40 in june mrs lees income = 60 * 110 / 100 = 66 in june total income = 66 + 40 = 106 % of mrs lee ' s income = 66 / 106 = 62.26 ( a )"
a = 100 + 10 b = 100 + 10 c = b + 100 d = a / c e = d * 100
a ) 10 mps , b ) 16.67 mps , c ) 97.67 mps , d ) 17.67 mps , e ) 18 mps
b
multiply(const_0_2778, 60)
express a speed of 60 kmph in meters per second ?
"60 * 5 / 18 = 16.67 mps answer : b"
a = const_0_2778 * 60
a ) 40 , b ) 80 , c ) 50 , d ) 60 , e ) 50
c
subtract(100, 50)
a person decided to build a house in 100 days . he employed 100 men in the beginning and 100 more after 50 days and completed the construction in stipulated time . if he had not employed the additional men , how many days behind schedule would it have been finished ?
200 men do the rest of the work in 100 - 50 = 50 days 100 men can do the rest of the work in 50 * 200 / 100 = 100 days required number of days = 100 - 50 = 50 days answer is c
a = 100 - 50
a ) 18 , b ) 16 , c ) 17 , d ) 19 , e ) 15
d
add(divide(subtract(140, 30), 6), const_1)
how many multiples of 6 are there between 30 and 140 , 30 and 140 inclusive ?
"6 multiples are . . . 30 , 36,42 , 48,54 , 60,66 , 72,78 , 84,90 , 96,102 , 108,114 , 120,126 , 132,138 , . . . , . . . , the answer is = 19 answer is d"
a = 140 - 30 b = a / 6 c = b + 1
a ) 25 , b ) 30 , c ) 40 , d ) 45 , e ) 36
e
divide(400, multiply(const_0_2778, subtract(46, 6)))
how many seconds will a 400 metre long train take to cross a man running with a speed of 6 km / hr in the direction of the moving train if the speed of the train is 46 km / hr ?
"explanation : speed of train relatively to man = ( 46 - 6 ) km / hr = 40 km / hr = ( 40 x 5 / 18 ) m / sec = 11.11 m / sec time taken to pass the man = ( 400 / 11.11 ) sec = 36 sec . answer : e"
a = 46 - 6 b = const_0_2778 * a c = 400 / b
a ) 20 , b ) 23 , c ) 25 , d ) 32 , e ) 42
d
divide(add(24, 40), const_2)
if x + y = 24 , x - y = 40 , for integers of x and y , x = ?
"x + y = 24 x - y = 40 2 x = 64 x = 32 answer is d"
a = 24 + 40 b = a / 2
a ) 60 , b ) 72 , c ) 72.5 , d ) 73 , e ) 73.5
a
divide(1, divide(add(multiply(const_3600, divide(1, 75)), 12), const_3600))
a car traveling at a certain constant speed takes 12 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 75 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ?
"many approaches are possible , one of them : let the distance be 1 kilometer . time to cover this distance at 75 kilometers per hour is 1 / 75 hours = 3,600 / 75 seconds = 48 seconds ; time to cover this distance at regular speed is 48 + 12 = 60 seconds = 60 / 3,600 hours = 1 / 60 hours ; so , we get that to cover 1 kilometer 1 / 60 hours is needed - - > regular speed 60 kilometers per hour ( rate is a reciprocal of time or rate = distance / time ) . answer : a"
a = 1 / 75 b = 3600 * a c = b + 12 d = c / 3600 e = 1 / d
a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50
a
multiply(add(const_2, const_3), 6)
if 222 = 6 333 = 12 444 = 20 then , 555 = ? ?
222 = 2 * 2 + 2 = 6 333 = 3 * 3 + 3 = 12 444 = 4 * 4 + 4 = 20 555 = 5 * 5 + 5 = 30 answer : a
a = 2 + 3 b = a * 6
a ) 0 , b ) 1 , c ) 4 , d ) 6 , e ) 8
a
divide(log(6), log(power(6, 11)))
if n = 6 ^ 11 – 6 , what is the units digit of n ?
always divide the power ( incase 11 ) by 4 and use the remainder as the new power . the question now becomes 6 ^ 3 - 6 . now 6 ^ 3 has last digit 6 . , we subtract 6 from 6 = 0 is the answer . option a
a = math.log(6) b = 6 ** 11 c = math.log(b) d = a / c
a ) 205 , b ) 200 , c ) 180 , d ) 195 , e ) 175
d
divide(25, 0.128)
a certain industrial loom weaves 0.128 meters of cloth every second . approximately how many seconds will it take for the loom to weave 25 meters of cloth ?
"explanation : let the required number of seconds be x more cloth , more time , ( direct proportion ) hence we can write as ( cloth ) 0.128 : 25 : : 1 : x β‡’ 0.128 x = 25 β‡’ x = 25 / 0.128 = 25000 / 128 = 3125 / 16 β‰ˆ 195 answer : option d"
a = 25 / 0
a ) 11.6 kmph , b ) 12.9 kmph , c ) 12.8 kmph , d ) 11.9 kmph , e ) 12.6 kmph
a
divide(add(14, 10), const_2)
a man goes from a to b at a speed of 14 kmph and comes back to a at a speed of 10 kmph . find his average speed for the entire journey ?
"distance from a and b be ' d ' average speed = total distance / total time average speed = ( 2 d ) / [ ( d / 14 ) + ( d / 10 ] = ( 2 d ) / [ 12 d / 70 ) = > 11.6 kmph . answer : a"
a = 14 + 10 b = a / 2
a ) 101 , b ) 107 , c ) 108 , d ) 109 , e ) 110
a
add(80, divide(80, 20))
p software has coding line 5 % more than n , n software has coding line 4 / 20 more than m . m software has 80 lines of coding . find p lines .
"m s / w has 80 line of code n s / w has = 80 + 80 * 4 / 20 = 96 line of code p s / w 5 % more n ' code 96 + 4.8 = 100.8 or 101 line of code answer : a"
a = 80 / 20 b = 80 + a
a ) 1 , b ) 8 , c ) 6 , d ) 4 , e ) 3
e
subtract(5, multiply(divide(subtract(3, 1), 2), 2))
if f ( x ) = 5 - 2 x , then f - 1 ( - 3 ) = ?
find f - 1 ( x ) and then find f - 1 ( - 3 ) y = 5 - 2 x , given x = 5 - 2 y , interchange x and y 2 y = 5 - x , y = log 2 ( 5 - x ) , solve for y f - 1 ( x ) = log 2 ( 5 - x ) , inverse function f - 1 ( - 3 ) = log 2 ( 5 + 3 ) = log 2 ( 23 ) = 3 correct answer e
a = 3 - 1 b = a / 2 c = b * 2 d = 5 - c
a ) 3 : 5 , b ) 5 : 7 , c ) 7 : 9 , d ) 9 : 11 , e ) 11 : 13
d
divide(add(1, divide(7, 2)), add(divide(7, 2), divide(4, 2)))
the weight of every type a widget is the same , the weight of every type b widget is the same , and the weight of every type c widget is the same . if the weight of 7 type a widgets is equal to the weight of 2 type b widgets , and the weight of 4 type b widgets is equal to the weight of 7 type c widgets . what is the ratio of the total weight of 1 type a widget and 1 type b widget , to the total weight of 1 type b widget and 1 type c widget ?
"4 b = 7 c and so b = 7 c / 4 7 a = 2 b and so a = 2 b / 7 = c / 2 a + b = c / 2 + 7 c / 4 = 9 c / 4 b + c = 7 c / 4 + c = 11 c / 4 the ratio of a + b : b + c = 9 : 11 the answer is d ."
a = 7 / 2 b = 1 + a c = 7 / 2 d = 4 / 2 e = c + d f = b / e
a ) 2902 , b ) 14000 , c ) 8402 , d ) 16800 , e ) none of these
b
divide(add(500, 200), 4)
( 500 + 200 ) Γ— 4 Γ— ( 3 + 2 ) = ?
"( 500 + 200 ) Γ— 4 Γ— ( 3 + 2 ) = ? or , ? = 700 Γ— 4 Γ— 5 = 14000 answer b"
a = 500 + 200 b = a / 4
a ) 48 , b ) 36 , c ) 42 , d ) 11 , e ) 18
c
divide(700, multiply(subtract(63, 3), const_0_2778))
how many seconds will a 700 meter long train moving with a speed of 63 km / hr take to cross a man walking with a speed of 3 km / hr in the direction of the train ?
"explanation : here distance d = 700 mts speed s = 63 - 3 = 60 kmph = 60 x 5 / 18 m / s time t = = 42 sec . answer : c"
a = 63 - 3 b = a * const_0_2778 c = 700 / b
a ) $ 150 , b ) $ 152 , c ) $ 154 , d ) $ 156 , e ) $ 158
e
subtract(400, divide(subtract(multiply(400, const_2), 74), const_3))
joan and karl each bought a telescopes and the sum of their purchases was $ 400 . if twice of what joan paid was $ 74 more than what karl paid , what did joan pay for her telescope ?
j + k = 400 so k = 400 - j 2 j = k + 74 2 j = ( 400 - j ) + 74 3 j = 474 j = 158 the answer is e .
a = 400 * 2 b = a - 74 c = b / 3 d = 400 - c
a ) 24 , b ) 34 , c ) 44 , d ) 54 , e ) 64
b
add(subtract(150, const_100), const_4)
how many odd prime numbers are there less than 150 ?
"odd prime number less than 150 : 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 , 73 , 79 , 83 , 89 , 97 , 101 , 103 , 107 , 109 , 113 , 127 , 131 , 137 , 139 , 149 there is 34 the odd prime number answer is b"
a = 150 - 100 b = a + 4
a ) 4.1 days , b ) 4.8 days , c ) 3.8 days , d ) 2.8 days , e ) 4.5 days
b
inverse(add(divide(const_1, 8), divide(const_1, multiply(6, const_2))))
a can do a piece of work in 8 days . he works at it for 4 days and then b finishes it in 6 days . in how many days can a and b together finish the work ?
4 / 8 + 6 / x = 1 = > x = 12 1 / 8 + 1 / 12 = 5 / 24 = > 4.8 days answer : b
a = 1 / 8 b = 6 * 2 c = 1 / b d = a + c e = 1/(d)
a ) 1 kmph , b ) 2 kmph , c ) 3 kmph , d ) 2.5 kmph , e ) 3.5 kmph
a
divide(subtract(multiply(divide(1, 12), const_60), multiply(divide(1, 20), const_60)), const_2)
a boat moves upstream at the rate of 1 km in 20 minutes and down stream 1 km in 12 minutes . then the speed of the current is :
rate upstream = ( 1 / 20 * 60 ) = 3 kmph rate dowm stream = 1 / 12 * 60 = 5 kmph rate of the current = Β½ ( 5 - 3 ) = 1 kmph answer : a
a = 1 / 12 b = a * const_60 c = 1 / 20 d = c * const_60 e = b - d f = e / 2
a ) $ 40 , b ) $ 160 , c ) $ 220 , d ) $ 240 , e ) $ 260
a
subtract(multiply(add(multiply(10, add(const_3, const_2)), const_2), 10), multiply(40, const_12))
a parking garage rents parking spaces for $ 10 per week or $ 40 per month . how much does a person save in a year by renting by the month rather than by the week ?
"10 $ per week ! an year has 52 weeks . annual charges per year = 52 * 10 = 520 $ 30 $ per month ! an year has 12 months . annual charges per year = 12 * 40 = 480 $ 520 - 480 = 40 ans a"
a = 3 + 2 b = 10 * a c = b + 2 d = c * 10 e = 40 * 12 f = d - e
a ) 263 , b ) 500 , c ) 388.9 , d ) 368 , e ) 339
c
divide(70, divide(multiply(4, add(4, divide(1, 2))), const_100))
what sum of money will produce rs . 70 as simple interest in 4 years at 4 1 / 2 percent ?
"70 = ( p * 4 * 9 / 2 ) / 100 p = 388.9 answer : c"
a = 1 / 2 b = 4 + a c = 4 * b d = c / 100 e = 70 / d
a ) 34 days , b ) 40 days , c ) 46 days , d ) 50 days , e ) none of these
a
multiply(divide(34, 34), 34)
in a dairy farm , 34 cows eat 34 bags of husk in 34 days . in how many days one cow will eat one bag of husk ?
"explanation : less cows , more days ( indirect proportion ) less bags , less days ( direct proportion ) [ cows 1 34 bags 34 1 ] : : 34 : x = > x βˆ— 34 βˆ— 1 = 34 βˆ— 1 βˆ— 34 = > x = 34 option a"
a = 34 / 34 b = a * 34
a ) 180 cm 2 , b ) 112 cm 2 , c ) 776 cm 2 , d ) 666 cm 2 , e ) 886 cm 2
a
divide(multiply(24, 15), const_2)
if the sides of a triangle are 28 cm , 24 cm and 15 cm , what is its area ?
"the triangle with sides 28 cm , 24 cm and 15 cm is right angled , where the hypotenuse is 28 cm . area of the triangle = 1 / 2 * 24 * 15 = 180 cm 2 answer : a"
a = 24 * 15 b = a / 2
a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 14
e
multiply(add(const_2, const_3), const_2)
if y is the smallest positive integer such that 3,150 multiplied by y is the square of an integer , then y must be
"3150 * y = a ^ 2 3 * 1050 * y = a ^ 2 3 * 5 * 210 * y = a ^ 2 3 * 5 * 7 * 3 * 2 * 5 * y = a ^ 2 2 * 3 ^ 2 * 5 ^ 2 * 7 * y = a ^ 2 concept : the factors of squared integer should occur in pair . so , the only numbers left are 2 * 7 hence 14 ; answer : e"
a = 2 + 3 b = a * 2
a ) 30 , b ) 35 , c ) 38 , d ) 40 , e ) 45
a
subtract(add(10, add(20, 5)), 5)
10 play kabadi , 20 play kho kho only , 5 play both gmaes . then how many in total ?
"10 play kabadi = > n ( a ) = 10 , 5 play both gmaes . = > n ( anb ) = 5 20 play kho kho only , = > n ( b ) = n ( b only ) + n ( anb ) = 20 + 5 = 25 total = > n ( aub ) = n ( a ) + n ( b ) - n ( anb ) = 10 + 25 - 5 = 30 answer : a"
a = 20 + 5 b = 10 + a c = b - 5
a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22
d
divide(add(sqrt(add(multiply(multiply(210, const_2), const_4), const_1)), const_1), const_2)
if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 210 games will be played during the tournament . what is the number of participants ?
"let n be the number of participants . the number of games is nc 2 = n * ( n - 1 ) / 2 = 210 n * ( n - 1 ) = 420 = 21 * 20 ( trial and error ) the answer is d ."
a = 210 * 2 b = a * 4 c = b + 1 d = math.sqrt(c) e = d + 1 f = e / 2
a ) 24.6 m . , b ) 23.4 m . , c ) 22.5 m . , d ) 23.8 m . , e ) 25.5 m .
e
multiply(sqrt(divide(divide(324, 6), const_3)), const_3)
the length of a rectangular floor is more than its breadth by 500 % . if rs . 324 is required to paint the floor at the rate of rs . 6 per sq m , then what would be the length of the floor ?
"let the length and the breadth of the floor be l m and b m respectively . l = b + 500 % of b = l + 5 b = 6 b area of the floor = 324 / 6 = 54 sq m l b = 54 i . e . , l * l / 6 = 108 l 2 = 648 = > l = 25.5 m . answer : e"
a = 324 / 6 b = a / 3 c = math.sqrt(b) d = c * 3
a ) 42 , b ) 49 , c ) 56 , d ) 63 , e ) 70
b
multiply(divide(7, subtract(9, 7)), 14)
sandy is younger than molly by 14 years . if the ratio of their ages is 7 : 9 , how old is sandy ?
"let sandy ' s age be 7 x and let molly ' s age be 9 x . 9 x - 7 x = 14 x = 7 sandy is 49 years old . the answer is b ."
a = 9 - 7 b = 7 / a c = b * 14
a ) 100 , b ) 110 , c ) 151 , d ) 120 , e ) 140
a
multiply(multiply(1, const_1000), 1)
find the product of the localvalue and absolutevalue of 1 in 20168 ?
"local value of 1 = 1 x 100 = 100 place value of 1 = 1 there fore = 1 x 100 = 100 a"
a = 1 * 1000 b = a * 1
a ) βˆ’ 48 , b ) βˆ’ 2 , c ) 6 , d ) 9 , e ) 48
c
subtract(subtract(subtract(subtract(add(add(6, 7), subtract(6, 7)), const_1), const_1), const_1), const_1)
if a ( a - 6 ) = 7 and b ( b - 6 ) = 7 , where a β‰  b , then a + b =
"i . e . if a = - 1 then b = 7 or if a = 7 then b = - 1 but in each case a + b = - 1 + 7 = 6 answer : option c"
a = 6 + 7 b = 6 - 7 c = a + b d = c - 1 e = d - 1 f = e - 1 g = f - 1
a ) 51 , b ) 64 , c ) 11 , d ) 22 , e ) 33
e
multiply(const_3600, divide(divide(605, const_1000), add(60, 6)))
a train 605 meters long is running with a speed of 60 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ?
"speed of train relative to man = ( 60 + 6 ) km / hr = 66 km / hr [ 66 * 5 / 18 ] m / sec = [ 55 / 3 ] m / sec . time taken to pass the man = [ 605 * 3 / 55 ] sec = 33 sec answer : e"
a = 605 / 1000 b = 60 + 6 c = a / b d = 3600 * c
a ) 46 , b ) 48 , c ) 50 , d ) 52 , e ) 54
c
subtract(multiply(8, 61), subtract(multiply(13, 60), multiply(57, 6)))
the average of 13 numbers is 60 . average of the first 6 of them is 57 and that of the last 8 is 61 . find the 8 th number ?
"sum of all the 13 numbers = 13 * 60 = 780 sum of the first 6 of them = 6 * 57 = 342 sum of the last 8 of them = 8 * 61 = 488 so , the 8 th number = 342 + 488 - 780 = 50 . answer : c"
a = 8 * 61 b = 13 * 60 c = 57 * 6 d = b - c e = a - d
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11
b
multiply(divide(subtract(22, const_2), add(const_3, const_2)), const_2)
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 22 , then how old is b ?
"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 = 22 5 x = 20 = > x = 4 hence , b ' s age = 2 x = 8 years . answer : b"
a = 22 - 2 b = 3 + 2 c = a / b d = c * 2
a ) $ 341.50 , b ) $ 344.60 , c ) $ 347.70 , d ) $ 350.80 , e ) $ 353.90
c
add(add(multiply(multiply(add(divide(multiply(120, 65), const_100), 65), 2), 0.70), multiply(multiply(3, 65), 0.70)), multiply(2.20, add(3, 2)))
in a fuel station the service costs $ 2.20 per vehicle and every liter of fuel costs $ 0.70 . assuming that you fill up 3 mini - vans and 2 trucks , what will be the total cost , if a mini - van ' s tank is 65 liters and a truck ' s tank is 120 % bigger and they are all empty ?
"the service cost of 3 vans and 2 trucks is 5 * 2.20 $ 11.00 the fuel in 3 vans is 3 * 65 = 195 liters the fuel in 2 trucks is 2 * 65 * 2.2 = 286 liters the total fuel ( vans + trucks ) = 481 liters the total fuel cost is 481 * 0.7 = $ 336.70 the total cost is $ 336.70 + $ 11.00 = $ 347.70 the answer is c ."
a = 120 * 65 b = a / 100 c = b + 65 d = c * 2 e = d * 0 f = 3 * 65 g = f * 0 h = e + g i = 3 + 2 j = 2 * 20 k = h + j
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14
b
divide(10, divide(multiply(multiply(2, multiply(multiply(add(2, const_3), 2), const_3)), 2), 10))
four bells commence tolling together and toll at intervals of 2 , 3 , 4 and 5 seconds respectively . in 10 minutes , how many times do they toll together ?
"lcm of 2 , 3 , 4 , and 5 is 60 . so , after each 60 seconds , they would toll together . hence , in 10 minutes , they would toll 10 * 60 seconds / 60 seconds = 10 times but then the question says they commence tolling together . so , they basically also toll at thebeginning ( 0 second ) . so , total tolls together = 10 + 1 = 11 answer : b"
a = 2 + 3 b = a * 2 c = b * 3 d = 2 * c e = d * 2 f = e / 10 g = 10 / f
a ) 8.33 % , b ) 16 7 / 3 % , c ) 16 2 / 8 % , d ) 16 2 / 3 % , e ) 76 2 / 3 %
a
subtract(const_100, divide(multiply(550, const_100), 600))
an article is bought for rs . 600 and sold for rs . 550 , find the loss percent ?
"600 - - - - 50 100 - - - - ? = > 8.33 % answer : a"
a = 550 * 100 b = a / 600 c = 100 - b
a ) 12 , b ) 18 , c ) 20 , d ) 24 , e ) 36
e
divide(multiply(120, add(multiply(4, const_2), 4)), 40)
it takes printer a 4 more minutes more than printer b to print 40 pages . working together , the two printers can print 50 pages in 6 minutes . how long will it take printer a to print 120 pages ?
"if it takes 4 more minutes for a to print 40 pages than it takes b , it takes 5 more minutes for a to print 50 pages than it takes b . thus if b is the number of minutes than b takes to print 50 pages , we can write : 1 / b + 1 / ( b + 5 ) = 1 / 6 ( since in 1 minute , they print 1 / 6 th of the 50 page job ) 6 ( 2 b + 5 ) = b ( b + 5 ) b ^ 2 - 7 b - 30 = 0 ( b - 10 ) ( b + 3 ) = 0 b = 10 thus it takes a 15 minutes to print 50 pages and 15 * 120 / 50 = 36 minutes to print 120 pages ( answer e )"
a = 4 * 2 b = a + 4 c = 120 * b d = c / 40
a ) 20 % , b ) 30 % , c ) 40 % , d ) 70 % , e ) 80 %
d
subtract(100, 30)
john want to buy a $ 100 trouser at the store , but he think it ’ s too expensive . finally , it goes on sale for $ 30 . what is the percent decrease ?
"the is always the difference between our starting and ending points . in this case , it ’ s 100 – 30 = 70 . the β€œ original ” is our starting point ; in this case , it ’ s 100 . ( 70 / 100 ) * 100 = ( 0.7 ) * 100 = 70 % . d"
a = 100 - 30
a ) 5 , b ) 3 , c ) 2 , d ) 4 , e ) 0
d
subtract(power(3, 2), 5)
for the symbol , m ” n = n ^ 2 βˆ’ m for all values of m and n . what is the value of 5 ” 3 ?
"5 ” 3 = 9 - 5 = 4 answer : d"
a = 3 ** 2 b = a - 5
a ) 39 kmh , b ) 40 kmh , c ) 42 kmh , d ) 44 kmh , e ) 50 kmh
a
divide(const_3, add(add(divide(const_1, 80), divide(const_1, 24)), divide(const_1, 44)))
if a car went the first third of the distance at 80 kmh , the second third at 24 kmh , and the last third at 44 kmh , what was the average speed of the car for the entire trip ?
"assume d / 3 = 2640 ( this number is convenient because it is divisible by 80 , 24 and 44 ) so : 2640 = 80 * t 1 = 33 hrs 2640 = 24 * t 2 = 110 hrs 2640 = 44 * t 3 = 60 hrs t = t 1 + t 2 + t 3 = 203 hrs d = rt ( 2640 * 3 ) = r * 203 r = 39 answer : a"
a = 1 / 80 b = 1 / 24 c = a + b d = 1 / 44 e = c + d f = 3 / e
a ) 31 , b ) 32 , c ) 35 , d ) 30 , e ) 45
d
divide(add(408, 312), multiply(multiply(multiply(const_2, const_2), const_2), const_3))
there are 408 boys and 312 girls in a school which are to be divided into equal sections of either boys or girls alone . find the total number of sections thus formed .
"explanation : hcf ( 408 , 312 ) = 24 the number of boys or girls that can be placed in a section = 24 . thus the total number of sections is given by 408 / 24 + 312 / 24 = 17 + 13 = 30 answer : d"
a = 408 + 312 b = 2 * 2 c = b * 2 d = c * 3 e = a / d
a ) 35 yrs , b ) 40 yrs , c ) 50 yrs , d ) 60 yrs , e ) 70 yrs
e
add(add(multiply(const_2, 20), 20), 10)
in 20 years , a will be twice as old as b was 20 years ago . if a is now 10 years older than b , the present age of b is ?
"let b ' s present age = x years then a ' s present age = x + 10 years x + 10 + 20 = 2 ( x - 20 ) x + 30 = 2 x - 40 x = 70 years answer is e"
a = 2 * 20 b = a + 20 c = b + 10
a ) 5 kmph , b ) 4.95 kmph , c ) 6 kmph , d ) 7 kmph , e ) 8 kmph
b
divide(add(divide(44, 8), divide(44, 10)), 8)
a boat running downstream covers a distance of 44 km in 8 hours while for covering the same distance upstream , it takes 10 hours . what is the speed of the boat in still water ?
"explanation : speed downstream = 44 / 8 = 5.5 kmph speed upstream = 44 / 10 = 4.4 kmph speed of the boat in still water = ( 5.5 + 4.4 ) / 2 = 4.95 kmph answer : option b"
a = 44 / 8 b = 44 / 10 c = a + b d = c / 8
a ) $ 945 , b ) $ 965 , c ) $ 1025 , d ) $ 1050 , e ) $ 1105
a
subtract(multiply(105, divide(const_100, 10)), 105)
if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 105 , how much more remains to be paid ?
"90 % remains to be paid so the remaining amount is 9 * 105 = $ 945 . the answer is a ."
a = 100 / 10 b = 105 * a c = b - 105
['a ) 0.125', 'b ) 0.25', 'c ) 0.5', 'd ) 0.75', 'e ) not enough information to determine the rate']
e
multiply(square_area(cube_edge_by_volume(840)), 0.5)
the volume v of a rectangular swimming pool is 840 cubic meters and water is flowing into the swimming pool . if the surface level of the water is rising at the rate of 0.5 meters per minute , what is the rate , in cubic meters per minutes , at which the water is flowing into the swimming pool ?
the correct answer is e . there are not enough info to answer the question . a 840 cubic meters rectangle is built from : height * length * width . from the question we know the volume of the pool and the filling rate . a pool can have a height of 10 * width 8.4 * length 10 and have a volume of 840 cubic meters , and it can have a height of 1 meter , width of 100 meters and length of 8.4 . in both cases the pool will fill up in a different rate = e
a = square_area * (
a ) 6.32 , b ) 5.32 , c ) 7.32 , d ) 2.32 , e ) 1.32
a
divide(200, divide(2500, 79))
calculate the dividend from caleb ' s stock , if caleb he invested rs . 2500 at 79 to obtain an income of rs . 200 .
by investing rs . 2500 , income = rs . 200 by investing rs . 79 , income = 200 Γ£ β€” 79 / 2500 = 6.32 ie , dividend = 6.32 % answer is a .
a = 2500 / 79 b = 200 / a
a ) 100 , b ) 200 , c ) 500 , d ) 1000 , e ) 5000
c
divide(multiply(5.4, multiply(multiply(const_100, const_100), const_100)), add(multiply(const_100, const_100), multiply(multiply(const_4, const_2), const_100)))
in the biology lab of ` ` jefferson ' ' high school there are 5.4 * 10 ^ 6 germs , equally divided among 10,800 petri dishes . how many germs live happily in a single dish ?
5.4 * 10 ^ 6 / 10800 = = > 54 * 10 ^ 3 / 108 = = > 54000 / 108 = 500 answer : c
a = 100 * 100 b = a * 100 c = 5 * 4 d = 100 * 100 e = 4 * 2 f = e * 100 g = d + f h = c / g
a ) 12.5 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) 100 %
c
multiply(divide(subtract(divide(multiply(const_10, const_4), multiply(divide(subtract(const_100, 20), const_100), const_10)), const_4), const_4), const_100)
a part - time employee whose hourly wage was decreased by 20 percent decided to increase the number of hours worked per week so that the employee ' s total income did not change . by what percent r should the number of hours worked be increased ?
"correct answer : c solution : c . we can set up equations for income before and after the wage reduction . initially , the employee earns w wage and works h hours per week . after the reduction , the employee earns . 8 w wage and works x hours . by setting these equations equal to each other , we can determine the increase in hours worked : wh = . 8 wx ( divide both sides by . 8 w ) 1.25 h = x we know that the new number of hours worked r will be 25 % greater than the original number . the answer is c ."
a = 10 * 4 b = 100 - 20 c = b / 100 d = c * 10 e = a / d f = e - 4 g = f / 4 h = g * 100
a ) 2.45 % , b ) 5.62 % , c ) 5.35 % , d ) 6.9 % , e ) 6.25 %
e
divide(const_100, 16)
at what rate percent per annum will a sum of money double in 16 years .
"let principal = p then s . i . = p t = 16 yrs rate = ( 100 * p ) / ( p * 16 ) = 25 / 4 = 6.25 % p . a . answer is e"
a = 100 / 16