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in a class of 40 students , 2 students did not borrow any books from the library , 12 students each borrowed 1 book , 12 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed ?
"the class borrowed a total of 40 * 2 = 80 books . the 26 students who borrowed 0 , 1 , or 2 books borrowed a total of 12 + 12 * 2 = 36 . to maximize the number of books borrowed by 1 student , let ' s assume that 13 students borrowed 3 books and 1 student borrowed the rest . 80 - 36 - 3 * 13 = 5 the maximum number of books borrowed by any student is 5 . the answer is c ."
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
c
subtract(multiply(40, 2), add(multiply(subtract(subtract(40, add(add(multiply(12, 1), 12), 2)), 1), 3), add(multiply(12, 1), multiply(12, 2))))
multiply(n0,n1)|multiply(n2,n3)|multiply(n1,n4)|add(#1,#2)|add(n4,#1)|add(n1,#4)|subtract(n0,#5)|subtract(#6,n3)|multiply(n6,#7)|add(#3,#8)|subtract(#0,#9)|
general
shop offered 40 % offer for every shirt , smith bought a shirt at rs . 560 . and what was the shop ' s original selling price ?
"sp * ( 60 / 100 ) = 560 sp = 9.33 * 100 = > cp = 933 answer : d"
a ) 500 , b ) 550 , c ) 600 , d ) 933 , e ) 750
d
divide(560, subtract(const_1, divide(40, const_100)))
divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)|
gain
the average score of a cricketer for 12 matches is 48 runs . if the average for first 8 matches is 40 , then average for last 4 matches is
explanation : = ( 48 × 12 ) − ( 40 × 8 ) / 4 = ( 576 − 320 ) / 4 = 64 answer : option c
a ) 33.25 , b ) 89 , c ) 64 , d ) 68 , e ) 90
c
subtract(add(add(48, 12), divide(40, 8)), subtract(4, const_3))
add(n0,n1)|divide(n3,n2)|subtract(n4,const_3)|add(#0,#1)|subtract(#3,#2)
general
the average of 40 results is 30 and the average of other 30 results is 40 . what is the average of all the results ?
answer sum of 70 result = sum of 40 result + sum of 30 result . = 40 x 30 + 30 x 40 = 2400 / 70 correct option : a
a ) 34 , b ) 25 , c ) 48 , d ) 50 , e ) none
a
divide(add(multiply(40, 30), multiply(30, 40)), add(40, 30))
add(n0,n1)|multiply(n0,n1)|add(#1,#1)|divide(#2,#0)
general
a candidate got 35 % of the votes polled and he lost to his rival by 2340 votes . how many votes were cast ?
"35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % - - - - - - - - - - 2340 100 % - - - - - - - - - ? = > 7800 answer : d"
a ) 7500 , b ) 3388 , c ) 2665 , d ) 7800 , e ) 2661
d
divide(2340, subtract(subtract(const_1, divide(35, const_100)), divide(35, const_100)))
divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)|
gain
a and b started a business in partnership investing rs . 20,000 and rs . 15,000 respectively . after 6 months , c joined them with rs . 20,000 . whatwill be b ' s share in total profit of rs . 25,000 earned at the end of 2 years from the startingof the business ?
"a : b : c = ( 20,000 x 24 ) : ( 15,000 x 24 ) : ( 20,000 x 18 ) = 4 : 3 : 3 . b ' s share = rs . 25000 x 3 / 10 = rs . 7,500 . e"
a ) rs . 5,000 , b ) rs . 5,500 , c ) rs . 5,700 , d ) rs . 6,500 , e ) rs . 7,500
e
multiply(multiply(add(const_4, const_1), const_4), multiply(2, multiply(const_3, const_4)))
add(const_1,const_4)|multiply(const_3,const_4)|multiply(#0,const_4)|multiply(n5,#1)|multiply(#2,#3)|
gain
in a box of 12 pens , a total of 4 are defective . if a customer buys 2 pens selected at random from the box , what is the probability that neither pen will be defective ?
"method - 1 there are 9 fine pieces of pen and 4 defective in a lot of 12 pens i . e . probability of first pen not being defective = ( 8 / 12 ) i . e . probability of second pen not being defective = ( 7 / 11 ) [ 11 pen remaining with 8 defective remaining considering that first was defective ] probability of both pen being non - defective = ( 8 / 12 ) * ( 7 / 11 ) = 14 / 33 answer : option c"
a ) 1 / 6 , b ) 2 / 33 , c ) 14 / 33 , d ) 9 / 16 , e ) 3 / 44
c
multiply(divide(subtract(12, 4), 12), divide(subtract(subtract(12, 4), const_1), subtract(12, const_1)))
subtract(n0,n1)|subtract(n0,const_1)|divide(#0,n0)|subtract(#0,const_1)|divide(#3,#1)|multiply(#2,#4)|
general
a bowl was filled with 10 ounces of water , and 0.007 ounce of the water evaporated each day during a 50 - day period . what percent of the original amount of water evaporated during this period ?
"total amount of water evaporated each day during a 50 - day period = . 007 * 50 = . 007 * 100 / 2 = . 7 / 2 = . 35 percent of the original amount of water evaporated during this period = ( . 35 / 10 ) * 100 % = 3.5 % answer e"
a ) 0.004 % , b ) 0.04 % , c ) 0.40 % , d ) 4 % , e ) 3.5 %
e
multiply(divide(multiply(50, 0.007), 10), const_100)
multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100)|
gain
the price of a certain painting increased by 30 % during the first year and decreased by 20 % during the second year . the price of the painting at the end of the 2 - year period was what percent of the original price ?
"easiest thing to do : assume that price is 100 price at the end of yr 1 : 100 + 30 = 130 price at the end of year 2 = 130 - 130 * 0.20 = 130 * 0.80 = 104 hence required answer = ( 104 / 100 ) * 100 % = 104 % answer is e ."
a ) 102 % , b ) 105 % , c ) 120 % , d ) 135 % , e ) 104 %
e
add(subtract(subtract(30, 20), divide(multiply(20, 30), const_100)), const_100)
multiply(n0,n1)|subtract(n0,n1)|divide(#0,const_100)|subtract(#1,#2)|add(#3,const_100)|
general
the least whole number which when subtracted from both the terms of the ratio 6 : 7 to give a ra ɵ o less than 16 : 21 , is
explanation : let x is subtracted . then , ( 6 − x ) ( 7 − x ) < 162121 ( 6 � x ) < 16 ( 7 � x ) = > 5 x > 14 = x > 2.8 least such number is 3 answer : a
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
a
add(floor(divide(subtract(multiply(21, 6), multiply(7, 16)), subtract(21, 16))), const_1)
multiply(n0,n3)|multiply(n1,n2)|subtract(n3,n2)|subtract(#0,#1)|divide(#3,#2)|floor(#4)|add(#5,const_1)
other
in what time will a train 134 m long cross an electric pole , it its speed be 169 km / hr ?
"speed = 169 * 5 / 18 = 46.9 m / sec time taken = 134 / 46.9 = 2.9 sec . answer : c"
a ) 2.5 , b ) 2.7 , c ) 2.9 , d ) 2.3 , e ) 2.1
c
divide(134, multiply(169, const_0_2778))
multiply(n1,const_0_2778)|divide(n0,#0)|
physics
a train 75 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ?
"speed of the train relative to man = ( 75 / 10 ) m / sec = ( 15 / 2 ) m / sec . [ ( 15 / 2 ) * ( 18 / 5 ) ] km / hr = 27 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 27 = = > x = 32 km / hr . answer : e"
a ) 28 , b ) 50 , c ) 88 , d ) 22 , e ) 32
e
divide(divide(subtract(75, multiply(multiply(10, const_0_2778), 10)), 5), const_0_2778)
multiply(n2,const_0_2778)|multiply(n2,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)|
physics
a train 200 m long crosses a platform 150 m long in 20 sec ; find the speed of the train ?
"d = 200 + 150 = 350 t = 20 s = 350 / 20 * 18 / 5 = 63 kmph . answer : b"
a ) 73 , b ) 63 , c ) 65 , d ) 70 , e ) 83
b
subtract(multiply(20, multiply(150, const_0_2778)), 200)
multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)|
physics
if ( 5 ^ 13 ) ( 9 ^ 7 ) = 3 ( 15 ^ x ) , what is the value of x ?
( 5 ^ 13 ) ( 9 ^ 7 ) = 3 ( 15 ^ x ) = > 5 ^ 13 * 3 ^ 14 = 3 * 3 ^ x * 5 ^ x = > 5 ^ 13 * 3 ^ 14 = 3 ^ ( x + 1 ) * 5 ^ x value of x = 13 answer d
a ) 7 , b ) 9 , c ) 11 , d ) 13 , e ) 15
d
divide(log(divide(multiply(power(5, 13), power(9, 7)), 3)), log(15))
log(n5)|power(n0,n1)|power(n2,n3)|multiply(#1,#2)|divide(#3,n4)|log(#4)|divide(#5,#0)
general
in one hour , a boat goes 19 km / hr along the stream and 11 km / hr against the stream . the speed of the boat in still water ( in km / hr ) is :
"explanation : let the speed downstream be a km / hr and the speed upstream be b km / hr , then speed in still water = 1 / 2 ( a + b ) km / hr rate of stream = 1 / 2 ( a − b ) km / hr speed in still water = 1 / 2 ( 19 + 11 ) kmph = 15 kmph . answer : option d"
a ) 12 kmph , b ) 13 kmph , c ) 14 kmph , d ) 15 kmph , e ) 16 kmph
d
stream_speed(19, 11)
stream_speed(n0,n1)|
physics
a , b , c , d , e , f , g , h sitting in a row what is the probability that a , b , d are sitting together ?
"total number of arrangement is = 8 ! = 40320 favorable event i . e a , b and d can be arranged in 3 ! and the remaining can be arranged in 6 ! since abd can be in any of the six positions . so 3 ! * 6 ! / 8 ! = 3 / 28 answer : b"
a ) 1 / 5 , b ) 3 / 28 , c ) 1 / 7 , d ) 1 / 14 , e ) 5 / 28
b
divide(multiply(factorial(add(subtract(add(const_4, const_4), const_3), const_1)), factorial(const_3)), factorial(add(const_4, const_4)))
add(const_4,const_4)|factorial(const_3)|factorial(#0)|subtract(#0,const_3)|add(#3,const_1)|factorial(#4)|multiply(#5,#1)|divide(#6,#2)|
probability
the divisor is 21 , the quotient is 14 and the remainder is 7 . what is the dividend ?
d = d * q + r d = 21 * 14 + 7 d = 294 + 7 d = 301
a ) 201 , b ) 394 , c ) 302 , d ) 301 , e ) 294
d
add(multiply(21, 14), 7)
multiply(n0,n1)|add(n2,#0)
general
tabby is training for a triathlon . she swims at a speed of 1 mile per hour . she runs at a speed of 6 miles per hour . she wants to figure out her average speed for these two events . what is the correct answer for her ?
"( 1 mph + 6 mph ) / 2 = 3.5 mph correct option is : c"
a ) 8 mph , b ) 5.25 mph , c ) 3.5 mph , d ) 4 mph , e ) 0.5 mph
c
divide(add(1, 6), const_2)
add(n0,n1)|divide(#0,const_2)|
physics
what annual payment will discharge a debt of rs . 1050 due in 2 years at the rate of 5 % compound interest ?
"explanation : let each installment be rs . x . then , x / ( 1 + 5 / 100 ) + x / ( 1 + 5 / 100 ) 2 = 1050 820 x + 1050 * 441 x = 564.69 so , value of each installment = rs . 564.69 answer : option e"
a ) 993.2 , b ) 551.25 , c ) 534.33 , d ) 543.33 , e ) 564.69
e
divide(multiply(power(add(divide(5, const_100), const_1), 2), 1050), 2)
divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|divide(#3,n1)|
gain
a dishonest dealer professes to sell goods at the cost price but uses a weight of 880 grams per kg , what is his percent ?
"explanation : 880 - - - 120 100 - - - ? = > 13.6 % answer : b"
a ) 15 % , b ) 13.6 % , c ) 65 % , d ) 45 % , e ) 35 %
b
subtract(multiply(divide(const_100, 880), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100)
add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100)|
gain
the number of positive integer solutions for the equation x + y + z + t = 21 is
"the number of positive integer solutions for the equatio fx 1 + x 2 + ⋯ + xn = k ( k - 1 ) c ( n - 1 ) - where k is the number and n is number of variable in the equation . 21 - 1 c 4 - 1 = 20 c 3 = 1140 answer : a"
a ) 1140 , b ) 1145 , c ) 1240 , d ) 1340 , e ) 15640
a
divide(factorial(subtract(21, const_1)), multiply(factorial(subtract(const_4, const_1)), factorial(subtract(subtract(21, const_1), subtract(const_4, const_1)))))
subtract(n0,const_1)|subtract(const_4,const_1)|factorial(#0)|factorial(#1)|subtract(#0,#1)|factorial(#4)|multiply(#3,#5)|divide(#2,#6)|
general
a shopkeeper buys mangoes at the rate of 6 a rupee and sells them at 5 a rupee . find his net profit or loss percent ?
"the total number of mangoes bought by the shopkeeper be 30 . if he buys 6 a rupee , his cp = 5 he selling at 5 a rupee , his sp = 6 profit = sp - cp = 6 - 5 = 1 profit percent = 1 / 5 * 100 = 20 % answer : c"
a ) 34 % , b ) 33 % , c ) 20 % , d ) 35 % , e ) 30 %
c
divide(multiply(5, const_100), 6)
multiply(n1,const_100)|divide(#0,n0)|
gain
machine p and machine q are each used to manufacture 330 sprockets . it takes machine p 10 hours longer to produce 330 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ?
"p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 330 / x = 330 / 1.1 x + 10 1.1 ( 330 ) = 330 + 11 x 11 x = 33 x = 3 the answer is c ."
a ) 5 , b ) 15 , c ) 3 , d ) 95 , e ) 125
c
divide(subtract(330, divide(330, add(divide(10, const_100), const_1))), 10)
divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|divide(#3,n1)|
gain
at veridux corporation , there are 250 employees . of these , 90 are female , and the rest are males . there are a total of 40 managers , and the rest of the employees are associates . if there are a total of 160 male associates , how many female managers are there ?
250 employees : 90 male , 160 female 40 managers , 210 associates 160 male associates implies 50 female associates which means the remaining 40 females must be managers a . 40
a ) 40 , b ) 20 , c ) 25 , d ) 30 , e ) 35
a
multiply(40, const_1)
multiply(n2,const_1)
general
think of a number , divide it by 5 and add 8 to it . the result is 61 . what is the number thought of ?
"explanation : 61 - 6 = 53 53 x 5 = 265 answer : d"
a ) 24 , b ) 77 , c ) 297 , d ) 265 , e ) 29
d
multiply(subtract(61, 8), 5)
subtract(n2,n1)|multiply(n0,#0)|
general
the ratio by weight , measured in pounds , of books to clothes to electronics in a suitcase initially stands at 7 : 4 : 3 . someone removes 8 pounds of clothing from the suitcase , thereby doubling the ratio of books to clothes . how many pounds do the electronics in the suitcase weigh ?
the weights of the items in the suitcase are 7 k , 4 k , and 3 k . if removing 8 pounds of clothes doubles the ratio of books to clothes , then 8 pounds represents half the weight of the clothes . 2 k = 8 pounds and then k = 4 pounds . the electronics weigh 3 ( 4 ) = 12 pounds . the answer is d .
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14
d
multiply(3, divide(negate(multiply(add(add(7, 4), 3), 8)), subtract(multiply(add(add(7, 4), 3), const_2), multiply(add(add(7, 4), 3), 4))))
add(n0,n1)|add(n2,#0)|multiply(n3,#1)|multiply(#1,const_2)|multiply(n1,#1)|negate(#2)|subtract(#3,#4)|divide(#5,#6)|multiply(n2,#7)
other
the radius of a semi circle is 2.1 cm then its perimeter is ?
"36 / 7 r = 2.1 = 10.80 answer : d"
a ) 32.5 , b ) 11.12 , c ) 32.1 , d ) 10.8 , e ) 32.3
d
add(divide(circumface(2.1), const_2), multiply(2.1, const_2))
circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)|
physics
the dimensions of a rectangular solid are 4 inches , 5 inches , and 10 inches . if a cube , a side of which is equal to one of the dimensions of the rectangular solid , is placed entirely within thespherejust large enough to hold the cube , what the ratio of the volume of the cube to the volume within thespherethat is not occupied by the cube ?
answer : c .
a ) 10 : 17 , b ) 2 : 5 , c ) 10 : 19 , d ) 25 : 7 , e ) 32 : 25
c
divide(const_10, add(add(multiply(10, const_2), const_2), const_1))
multiply(n2,const_2)|add(#0,const_2)|add(#1,const_1)|divide(const_10,#2)|
geometry
which number should replace both the asterisks in ( * / 21 ) x ( * / 189 ) = 1 ?
"answer let ( y / 21 ) x ( y / 189 ) = 1 then , y 2 = 21 x 189 = 21 x 21 x 9 ∴ y = ( 21 x 3 ) = 63 . option : b"
a ) 21 , b ) 63 , c ) 3969 , d ) 147 , e ) 167
b
sqrt(multiply(189, 21))
multiply(n0,n1)|sqrt(#0)|
general
a tank contains 10,000 gallons of a solution that is 6 percent sodium chloride by volume . if 6,000 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ?
"the remaining solution will be approximately what percent sodium chloride ? means : what percent of the remaining solution is sodium chloride . now , since the remaining solution is 10,000 - 6,000 = 4,000 gallons and sodium chloride is 600 gallons ( 6 % of initial solution of 10,000 gallons ) then sodium chloride is 600 / 4,000 * 100 = 15 % of the remaining solution of 4,000 gallons . answer : e ."
a ) 1.25 % , b ) 3.75 % , c ) 6.25 % , d ) 6.67 % , e ) 15 %
e
multiply(divide(multiply(multiply(const_100, const_100), divide(6, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100)
add(const_2,const_3)|divide(n1,const_100)|multiply(const_100,const_100)|multiply(#1,#2)|multiply(#0,const_2)|multiply(#0,const_100)|multiply(#4,const_100)|multiply(#0,#6)|add(#7,#5)|subtract(#2,#8)|divide(#3,#9)|multiply(#10,const_100)|
gain
two pipes a and b can fill a cistern in 12 and 20 minutes respectively , and a third pipe c can empty it in 30 minutes . how long will it take to fill the cistern if all the three are opened at the same time ?
"1 / 12 + 1 / 20 - 1 / 30 = 1 / 10 10 / 1 = 10 answer : d"
a ) 15 min , b ) 13 min , c ) 12 min , d ) 10 min , e ) 17 min
d
add(multiply(12, subtract(const_1, multiply(add(inverse(12), inverse(20)), 30))), 30)
inverse(n0)|inverse(n1)|add(#0,#1)|multiply(#2,n2)|subtract(const_1,#3)|multiply(n0,#4)|add(n2,#5)|
physics
two tests had the same maximum mark . the pass percentages in the first and the second test were 40 % and 45 % respectively . a candidate scored 231 marks in the second test and failed by 48 marks in that test . find the pass mark in the first test ?
"let the maximum mark in each test be m . the candidate failed by 48 marks in the second test . pass mark in the second test = 231 + 48 = 279 45 / 100 m = 279 pass mark in the first test = 40 / 100 m = 40 / 45 * 279 = 248 . answer : e"
a ) 768 , b ) 243 , c ) 246 , d ) 625 , e ) 248
e
add(231, 48)
add(n2,n3)|
gain
trains a and b start simultaneously from stations 200 miles apart , and travel the same route toward each other on adjacent parallel tracks . if train a and train b travel at a constant rate of 20 miles per hour and 20 miles per hour , respectively , how many miles will train a have traveled when the trains pass each other , to the nearest mile ?
"since we know the distance ( 200 ) and the combined rate ( 40 ) , we plug it into the formula : distance = rate * time 200 = 40 * time we can solve for the time they will meet cause we added the rate of train a and train b together . so the time will be 200 / 40 from dividing 40 on both sides to isolate time in the equation above . time will be 5 hours so now you can plug that in for train a ’ s distance . distance = rate * time distance = 20 * 5 distance = 100 according to answer choice a ."
a ) 100 , b ) 133 , c ) 150 , d ) 167 , e ) 188
a
multiply(20, divide(200, add(20, 20)))
add(n1,n2)|divide(n0,#0)|multiply(n1,#1)|
physics
x = 13.165 y = 7.686 z = 11.545 the number a is obtained by first rounding the values of x , y , and z to the hundredths place and then adding the resulting values . the number b is obtained by first adding the values of x , y , and z and then rounding the sum to the hundredths place . what is the value of a – b ?
rounding off the values : x = 13.17 y = 7.69 z = 11.55 a = 13.17 + 7.69 + 11.55 = 32.41 x + y + z = 13.165 + 7.686 + 11.545 = 32.206 b = 32.21 a - b = 0.2 option e
a ) 0.01 , b ) 0.02 , c ) 0.03 , d ) 0.1 , e ) 0.2
e
divide(subtract(add(floor(multiply(add(add(13.165, 7.686), 11.545), const_100)), const_1), add(floor(multiply(11.545, const_100)), add(floor(multiply(13.165, const_100)), floor(multiply(7.686, const_100))))), const_10)
add(n0,n1)|multiply(n0,const_100)|multiply(n1,const_100)|multiply(n2,const_100)|add(n2,#0)|floor(#1)|floor(#2)|floor(#3)|add(#5,#6)|multiply(#4,const_100)|add(#8,#7)|floor(#9)|add(#11,const_1)|subtract(#12,#10)|divide(#13,const_10)
general
the area of a square is 144 m ² . find its perimeter .
area of square = side × side given ; area of square = 144 m ² therefore , side ² = 144 m ² therefore , side = √ ( 144 m ² ) = √ ( 2 × 2 × 2 × 2 × 3 × 3 ) m ² = 2 × 2 × 3 m = 12 m now , the perimeter of the square = 4 x side = 4 × 12 m = 48 m answer : e
['a ) 18 m', 'b ) 28 m', 'c ) 38 m', 'd ) 42 m', 'e ) 48 m']
e
square_perimeter(sqrt(144))
sqrt(n0)|square_perimeter(#0)
geometry
how is 1 % expressed as a decimal fraction ?
"1 / 100 = 0.01 answer : c"
a ) 0.1 , b ) 0.001 , c ) 0.01 , d ) 0.0001 , e ) 1
c
divide(1, const_100)
divide(n0,const_100)|
gain
salesperson a ' s compensation for any week is $ 350 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 a ' s total sales for that week . for what amount of total weekly sales would both salepeople earn the same compensation ?
"sometime , setting up an equation is an easy way to go with : 350 + 0.06 ( x - 1000 ) = 0.08 x x = 14,500 ans : d"
a ) $ 21,000 , b ) $ 18,000 , c ) $ 15,000 , d ) $ 14,500 , e ) $ 4,000
d
divide(add(divide(subtract(350, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100))), divide(subtract(350, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100)))), 1,000)
divide(n1,const_100)|divide(n3,const_100)|multiply(#0,n2)|subtract(#1,#0)|subtract(n0,#2)|divide(#4,#3)|add(#5,#5)|divide(#6,n2)|
general
a , b , c and d enter into partnership . a subscribes 1 / 3 of the capital b 1 / 4 , c 1 / 5 and d the rest . how much share did a get in a profit of rs . 2430 ?
"2430 * 1 / 3 = 810 option b"
a ) s . 800 , b ) s . 810 , c ) s . 820 , d ) s . 900 , e ) s . 920
b
multiply(2430, divide(1, 3))
divide(n0,n1)|multiply(n6,#0)|
general
a person walks at a speed of 4 km / hr and runs at a speed of 8 km / hr . how many hours will the person require to cover a distance of 8 km , if the person completes half of the distance by walking and the other half by running ?
"time = 4 / 4 + 4 / 8 = 12 / 8 = 1.5 hours the answer is b ."
a ) 0.5 , b ) 1.5 , c ) 2.5 , d ) 3.5 , e ) 4.5
b
add(divide(divide(8, const_2), 4), divide(divide(8, const_2), 8))
divide(n2,const_2)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|
physics
dacid obtained 45 , 35 , 52 , 47 and 55 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ?
average = ( 45 + 35 + 52 + 47 + 55 ) / 5 = 234 / 5 = 47 . answer : c
a ) 79 , b ) 99 , c ) 47 , d ) 89 , e ) 45
c
divide(add(add(add(add(45, 35), 52), 47), 55), divide(const_10, const_2))
add(n0,n1)|divide(const_10,const_2)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)
general
if anangletis defined as 1 percent of 1 degree , then how many anglets are there in a quarter circle ?
"1 degree * 1 / 100 = 1 anglet so 1 degree = 100 anglets = > 90 degrees = 9000 anglets answer - c"
a ) 0.36 , b ) 3.6 , c ) 9000 , d ) 3,600 , e ) 36,000
c
multiply(divide(1, add(const_2, const_4)), multiply(divide(const_3600, const_10), const_100))
add(const_2,const_4)|divide(const_3600,const_10)|divide(n0,#0)|multiply(#1,const_100)|multiply(#2,#3)|
geometry
the product of a and b is equal to 10 more than twice the sum of a and b . if b = 9 , what is the value of b - a ?
"ab = 10 + 2 ( a + b ) 9 a = 10 + 2 a + 18 7 a = 28 a = 4 b - a = 9 - 4 = 5 b is the answer"
a ) 2 , b ) 5 , c ) 7 , d ) 24 , e ) 35
b
subtract(9, divide(add(multiply(9, const_2), 10), subtract(9, const_2)))
multiply(n1,const_2)|subtract(n1,const_2)|add(n0,#0)|divide(#2,#1)|subtract(n1,#3)|
general
how many times digit 2 is used while writing numbers from 400 to 700 ?
"in 400 to 500 there are 50 two ' s in 500 to 600 there are 50 two ' s in 600 to 700 there are 50 two ' s so total is 150 two ' s correct option : a"
a ) 150 , b ) 190 , c ) 250 , d ) 80 , e ) 130
a
add(add(divide(subtract(700, 400), const_10), multiply(add(const_10, const_1), add(const_10, const_1))), multiply(2, const_2))
add(const_1,const_10)|multiply(n0,const_2)|subtract(n2,n1)|divide(#2,const_10)|multiply(#0,#0)|add(#3,#4)|add(#5,#1)|
general
lionel left his house and walked towards walt ' s house , 48 miles away . two hours later , walt left his house and ran towards lionel ' s house . if lionel ' s speed was 2 miles per hour and walt ' s 6 miles per hour , how many miles had lionel walked when he met walt ?
in the first 2 hours lionel at the rate of 2 miles per hour covered distance = rate * time = 2 * 2 = 4 miles . so , the distance between him and walt was 48 - 4 = 44 miles when walt left his house . now , their combined rate to cover this distance was 2 + 6 = 8 miles per hour , hence they will meet ( they will cover that distance ) in time = distance / rate = 44 / 8 = 5.5 hours . total time that lionel was walking is 2 + 5.5 = 7.5 hours , which means that he covered in that time interval distance = rate * time = 2 * 7.5 = 15 miles . answer : b .
a ) 12 , b ) 15 , c ) 20 , d ) 24 , e ) 28
b
multiply(2, add(divide(subtract(48, multiply(2, const_2)), add(6, 2)), const_2))
add(n1,n2)|multiply(n1,const_2)|subtract(n0,#1)|divide(#2,#0)|add(#3,const_2)|multiply(n1,#4)
physics
due to construction , the speed limit along an 10 - mile section of highway is reduced from 55 miles per hour to 15 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ?
"old time in minutes to cross 10 miles stretch = 10 * 60 / 55 = 10 * 12 / 11 = 10.9 new time in minutes to cross 10 miles stretch = 10 * 60 / 15 = 10 * 4 = 40 time difference = 29.1 ans : c"
a ) a ) 6.24 , b ) b ) 8 , c ) c ) 29.1 , d ) d ) 15 , e ) e ) 24
c
max(multiply(subtract(add(55, 10), const_1), subtract(divide(10, 15), divide(10, 55))), const_4)
add(n0,n1)|divide(n0,n2)|divide(n0,n1)|subtract(#0,const_1)|subtract(#1,#2)|multiply(#3,#4)|max(#5,const_4)|
physics
the ratio of the volumes of two cubes is 2744 : 3375 . what is the ratio of their total surface areas ?
"ratio of the sides = ³ √ 2744 : ³ √ 3375 = 14 : 15 ratio of surface areas = 142 : 152 = 71 : 76 answer : a"
a ) 71 : 76 , b ) 71 : 131 , c ) 71 : 145 , d ) 71 : 167 , e ) 71 : 113
a
power(divide(2744, 3375), divide(const_1, const_3))
divide(n0,n1)|divide(const_1,const_3)|power(#0,#1)|
geometry
a and b started a business investing rs . 90,000 and rs 20,000 respectively . in what ratio the profit earned after 2 years be divided between a and b respectively ?
"a : b = 90000 : 20000 = 90 : 20 = 18 : 4 = 9 : 2 answer : b"
a ) 3 : 2 , b ) 9 : 2 , c ) 18 : 20 , d ) 1 : 4 , e ) 18 : 4
b
divide(add(multiply(multiply(const_3, const_3), add(multiply(const_3, const_3), const_1)), 2), multiply(2, add(multiply(const_3, const_3), const_1)))
multiply(const_3,const_3)|add(#0,const_1)|multiply(#1,#0)|multiply(n2,#1)|add(n2,#2)|divide(#4,#3)|
gain
the compound interest on $ 2000 at 10 % per annum is $ 662 . the period ( in years ) is ?
a = p ( 1 + r / 100 ) ^ t 2662 = 2000 ( 1 + 10 / 100 ) ^ t ( 11 / 10 ) ^ t = 2662 / 2000 ( 11 / 10 ) ^ t = ( 11 / 10 ) ^ 3 t = 3 years answer is b
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
b
divide(2000, 662)
divide(n0,n2)
gain
if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 150 , how much more remains to be paid ?
"10 / 100 p = 150 > > p = 150 * 100 / 10 = 1500 1500 - 150 = 1350 answer : c"
a ) $ 880 , b ) $ 990 , c ) $ 1,350 , d ) $ 1,100 , e ) $ 1,210
c
subtract(multiply(150, divide(const_100, 10)), 150)
divide(const_100,n0)|multiply(n1,#0)|subtract(#1,n1)|
general
what is 995 * 995 ?
"if you take a base of 1000 then 995 is 5 less than 1000 to get the product of 995 x 995 write like this 995 - 5 ( as 5 less than base 1000 ) 995 - 5 now 5 x 5 = 25 and 995 - 5 = 990 so 995 x 995 = 990025 . . . ( bingo the answer is e . you can even have a shortcut . . . . . . 5 x 5 = 25 . . . only answer choice has last three digits as 25 . . so no need to calculate 995 - 5 . after you get 5 x 5 you can straight way pick answer choice e ."
a ) 974,169 , b ) 974,219 , c ) 974,549 , d ) 985,019 , e ) 990,025
e
circle_area(divide(995, multiply(const_2, const_pi)))
multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)|
general
a train 125 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 15 seconds . the speed of the train is ?
"speed of the train relative to man = ( 125 / 15 ) m / sec = ( 25 / 3 ) m / sec . [ ( 25 / 3 ) * ( 18 / 5 ) ] km / hr = 30 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 30 = = > x = 35 km / hr . answer : a"
a ) 35 , b ) 50 , c ) 88 , d ) 22 , e ) 12
a
divide(divide(subtract(125, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778)
multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)|
physics
an object thrown directly upward is at a height of h feet after t seconds , where h = - 15 ( t - 3 ) ^ 2 + 150 . at what height , in feet , is the object 2 seconds after it reaches its maximum height ?
we see that h will be a maximum h = 150 when t - 3 = 0 , that is when t = 3 . at t = 5 , h = - 15 ( 5 - 3 ) ^ 2 + 150 = - 15 ( 4 ) + 150 = 90 the answer is a .
a ) 90 , b ) 98 , c ) 106 , d ) 120 , e ) 136
a
add(multiply(negate(15), power(subtract(add(3, 2), 3), const_2)), 150)
add(n1,n2)|negate(n0)|subtract(#0,n1)|power(#2,const_2)|multiply(#1,#3)|add(n3,#4)
general
the average of 6 observations is 13 . a new observation is included and the new average is decreased by 1 . the seventh observation is ?
let seventh observation = x . then , according to the question we have = > ( 78 + x ) / 7 = 12 = > x = 6 . hence , the seventh observation is 6 . answer : d
a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 7
d
subtract(multiply(subtract(13, 1), add(6, 1)), multiply(13, 6))
add(n0,n2)|multiply(n0,n1)|subtract(n1,n2)|multiply(#0,#2)|subtract(#3,#1)
general
the vertex of a rectangle are ( 1 , 0 ) , ( 9 , 0 ) , ( 1 , 4 ) and ( 9 , 4 ) respectively . if line l passes through the origin and divided the rectangle into two identical quadrilaterals , what is the slope of line l ?
"if line l divides the rectangle into two identical quadrilaterals , then it must pass through the center ( 5 , 2 ) . the slope of a line passing through ( 0,0 ) and ( 5 , 2 ) is 2 / 5 . the answer is d ."
a ) 4 , b ) 2 , c ) 1 / 4 , d ) 2 / 5 , e ) 5 / 6
d
divide(1, divide(add(subtract(9, 1), 4), 4))
subtract(n2,n0)|add(#0,n5)|divide(#1,n5)|divide(n0,#2)|
general
a test has 100 questions . each question has 5 options , but only 1 option is correct . if test - takers mark the correct option , they are awarded 1 point . however , if an answer is incorrectly marked , the test - taker loses 0.25 points . no points are awarded or deducted if a question is not attempted . a certain group of test - takers attempted different numbers of questions , but each test - taker still received the same net score of 40 . what is the maximum possible number of such test - takers ?
"a correct answers get you 1 point , an incorrect answer gets you minus 1 / 4 point and a skipped question gets you 0 points . since there are 200 total questions , there are a variety of ways to get a total of 40 points . let c be the number of correct answers and let i be the number of incorrect answers . to get 40 points , a test taker must have at least 40 correct answers . then c = > 40 . for every correct question above 40 , the test taker has 4 incorrect answers . then , the i = 4 * ( c - 40 ) . also , i + c < = 100 . thus 5 c < = 260 and so c < = 52 . then 40 < = c < = 52 and c can have 13 possible values . the answer is e ."
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13
e
add(add(divide(subtract(100, 40), 5), 1), 1)
subtract(n0,n5)|divide(#0,n1)|add(#1,n2)|add(#2,n2)|
general
a women in her conversation said ` ` if u reverse my own age , in figures represent my husbands age . he is of course senior to me and difference between our age is one 0 ne - eleventh of their sum . what is the woman ' s and her husbands age ?
let man , s age be xy the woman ' s age is yx then 10 x + y - 10 y - x = 1 / 11 ( 10 x + y + 10 y + x ) 9 x - 9 y = x + y 8 x = 10 y x / y = 10 / 8 = 5 / 4 so xy = 54 and yx = 45 answer : a
a ) 4554 , b ) 4657 , c ) 8457 , d ) 3458 , e ) 5254
a
add(add(multiply(multiply(const_1000, const_1), const_4), divide(multiply(const_1000, const_1), const_2)), multiply(add(const_4, const_1), const_10))
add(const_1,const_4)|multiply(const_1,const_1000)|divide(#1,const_2)|multiply(#1,const_4)|multiply(#0,const_10)|add(#2,#3)|add(#5,#4)
general
if 4 men working 10 hours a day earn rs . 600 per week , then 9 men working 6 hours a day will earn how much per week ?
"explanation : ( men 4 : 9 ) : ( hrs / day 10 : 6 ) : : 600 : x hence 4 * 10 * x = 9 * 6 * 600 or x = 9 * 6 * 600 / 4 * 10 = 810 answer : c"
a ) rs 840 , b ) rs 320 , c ) rs 810 , d ) rs 680 , e ) none of these
c
multiply(divide(multiply(9, 6), multiply(4, 10)), 600)
multiply(n3,n4)|multiply(n0,n1)|divide(#0,#1)|multiply(n2,#2)|
physics
a dog breeder currently has 9 breeding dogs . 6 of the dogs have exactly 1 littermate , and 3 of the dogs have exactly 2 littermates . if 2 dogs are selected at random , what is the probability q that both selected dogs are not littermates ?
"we have three pairs of dogs for the 6 with exactly one littermate , and one triplet , with each having exactly two littermates . so , in fact there are two types of dogs : those with one littermate - say a , and the others with two littermates - b . work with probabilities : choosing two dogs , we can have either one dog of type b or none ( we can not have two dogs both of type b ) . the probability of choosing one dog of type b and one of type a is 3 / 9 * 6 / 8 * 2 = 1 / 2 ( the factor of 2 for the two possibilities ba and ab ) . the probability of choosing two dogs of type a which are not littermates is 6 / 9 * 4 / 8 = 1 / 3 ( choose one a , then another a which is n ' t the previous one ' s littermate ) . the required probability is 1 / 2 + 1 / 3 = 5 / 6 . find the probability for the complementary event : choose aa or bb . probability of choosing two dogs of type a who are littermates is 6 / 9 * 1 / 8 = 1 / 12 . probability of choosing two dogs of type b ( who necessarily are littermates ) is 3 / 9 * 2 / 8 q = 1 / 12 . again , we obtain 1 - ( 1 / 12 + 1 / 12 ) = 5 / 6 . answer : c"
a ) 1 / 6 , b ) 2 / 9 , c ) 5 / 6 , d ) 7 / 9 , e ) 8 / 9
c
divide(const_5, 6)
divide(const_5,n1)|
other
a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 20.25 , the number of the member is the group is :
"money collected = ( 20.25 x 100 ) paise = 2025 paise numbers of members = 2025 squareroot = 45 answer c"
a ) 57 , b ) 67 , c ) 45 , d ) 47 , e ) 97
c
sqrt(multiply(20.25, const_100))
multiply(n0,const_100)|sqrt(#0)|
general
a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 12 sec . what is the length of the platform ?
"speed = 300 / 12 = 25 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 25 = > x = 675 m . answer : e"
a ) 187 m , b ) 350 m , c ) 267 m , d ) 287 m , e ) 675 m
e
subtract(multiply(speed(300, 12), 39), 300)
speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0)|
physics
mr . karan borrowed a certain amount at 6 % per annum simple interest for 9 years . after 9 years , he returned rs . 8510 / - . find out the amount that he borrowed .
"explanation : let us assume mr . karan borrowed amount is rs . a . ( the principal ) by formula of simple interest , s . i . = prt / 100 where p = the principal , r = rate of interest as a % , t = time in years s . i . = ( p * 6 * 9 ) / 100 = 54 p / 100 amount = principal + s . i . 8510 = p + ( 54 p / 100 ) 8510 = ( 100 p + 54 p ) / 100 8510 = 154 p / 100 p = ( 8510 * 100 ) / 154 = rs . 5525.974 answer : e"
a ) s . 5266 , b ) s . 5269 , c ) s . 5228 , d ) s . 5218 , e ) s . 5526
e
divide(8510, add(const_1, divide(multiply(6, 9), const_100)))
multiply(n0,n1)|divide(#0,const_100)|add(#1,const_1)|divide(n3,#2)|
gain
if the sum of two numbers is 24 and the sum of their squares is 400 , then the product of the numbers is
"according to the given conditions x + y = 24 and x ^ 2 + y ^ 2 = 400 now ( x + y ) ^ 2 = x ^ 2 + y ^ 2 + 2 xy so 24 ^ 2 = 400 + 2 xy so xy = 176 / 2 = 88 answer : d"
a ) 40 , b ) 44 , c ) 80 , d ) 88 , e ) 48
d
divide(subtract(power(24, const_2), 400), const_2)
power(n0,const_2)|subtract(#0,n1)|divide(#1,const_2)|
general
if the area of a square with sides of length 6 centimeters is equal to the area of a rectangle with a width of 4 centimeters , what is the length of the rectangle , in centimeters ?
"let length of rectangle = l 6 ^ 2 = l * 4 = > l = 36 / 4 = 9 answer d"
a ) 4 , b ) 8 , c ) 12 , d ) 9 , e ) 18
d
divide(power(6, const_2), 4)
power(n0,const_2)|divide(#0,n1)|
geometry
how many integer values t are there for x such that 1 < 3 x + 5 < 17 ?
1 < 3 x + 5 < 17 = > - 4 < 3 x < 12 = > - 4 / 3 < x < 4 x can take integer values t - 10 , 1 , 2 , 3 answer d
a ) two , b ) three , c ) four , d ) five , e ) six
d
add(floor(subtract(divide(subtract(17, 5), 5), divide(subtract(1, 5), 5))), const_2)
subtract(n3,n2)|subtract(n0,n2)|divide(#0,n2)|divide(#1,n2)|subtract(#2,#3)|floor(#4)|add(#5,const_2)
general
the sum of 3 consecutive multiples of 3 is 117 . what is the largest number ?
let the numbers be 3 x , 3 x + 3 and 3 x + 6 . then , 3 x + ( 3 x + 3 ) + ( 3 x + 6 ) = 117 9 x = 108 x = 12 largest number = 3 x + 6 = 42 answer : d
a ) 45 , b ) 48 , c ) 51 , d ) 42 , e ) 54
d
add(divide(117, 3), 3)
divide(n2,n0)|add(n0,#0)
general
if the cost price of 81 chocolates is equal to the selling price of 45 chocolates , the gain percent is :
"explanation : solution : let c . p . of each chocolate be re . 1 . then , c . p . of 45 chocolates = rs . 45 ; s . p . of 45 chocolates = rs . 81 . . ' . gain % = 36 * 100 / 45 = 80 % answer : a"
a ) 80 % , b ) 85 % , c ) 82 % , d ) 70 % , e ) 75 %
a
divide(const_100, divide(45, subtract(81, 45)))
subtract(n0,n1)|divide(n1,#0)|divide(const_100,#1)|
gain
company z has 50 employees . if the number of employees having birthdays on wednesday is more than the number of employees having birthdays on any other day of the week , each of which have same number of birth - days , what is the minimum number of employees having birthdays on wednesday .
"say the number of people having birthdays on wednesday is x and the number of people having birthdays on each of the other 6 days is y . then x + 6 y = 50 . now , plug options for x . only a and e give an integer value for y . but only for e x > y as needed . answer : e ."
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 12
e
add(const_4, add(floor(divide(50, add(const_4, const_3))), const_1))
add(const_3,const_4)|divide(n0,#0)|floor(#1)|add(#2,const_1)|add(#3,const_4)|
general
the cost price of an article is 64 % of the marked price . calculate the gain percent after allowing a discount of 18 % ?
"explanation : let marked price = rs . 100 . then , c . p . = rs . 64 , s . p . = rs . 82 gain % = 18 / 64 * 100 = 28.125 % . answer : option d"
a ) 37.5 % , b ) 48 % , c ) 50.5 % , d ) 28.125 % , e ) 56 %
d
multiply(subtract(divide(subtract(const_100, 18), 64), const_1), const_100)
subtract(const_100,n1)|divide(#0,n0)|subtract(#1,const_1)|multiply(#2,const_100)|
gain
the average age of 39 students in a group is 14 years . when teacher ' s age is included to it , the average increases by one . what is the teacher ' s age in years ?
"explanation : age of the teacher = ( 40 * 15 - 39 * 14 ) years = 54 years . answer : d"
a ) 35 years , b ) 45 years , c ) 51 years , d ) 54 years , e ) none of these
d
add(39, const_1)
add(n0,const_1)|
general
how many times are the hands of a clock at right angle in 2 weeks ?
"in 1 day , they are at right angles 44 times . in 14 days , they are at right angles 616 times . answer : option a"
a ) 616 , b ) 611 , c ) 661 , d ) 116 , e ) 666
a
divide(multiply(multiply(multiply(const_12, const_2), 2), subtract(multiply(const_12, const_4), const_4)), multiply(const_12, const_2))
multiply(const_12,const_2)|multiply(const_12,const_4)|multiply(n0,#0)|subtract(#1,const_4)|multiply(#2,#3)|divide(#4,#0)|
geometry
the membership of a committee consists of 3 english teachers , 4 mathematics teachers , and 2 social studies teachers . if 2 committee members are to be selected at random to write the committee ’ s report , what is the probability that the two members selected will both be maths teachers ?
probability of first member an english teacher = 3 / 9 probability of second member an english teacher = 2 / 8 probability of both being english teacher = 3 / 9 x 2 / 8 = 1 / 12 ( e )
a ) 2 / 3 , b ) 1 / 3 , c ) 2 / 9 , d ) 1 / 24 , e ) 1 / 12
e
multiply(divide(3, add(add(3, 4), 2)), divide(2, subtract(add(add(3, 4), 2), const_1)))
add(n0,n1)|add(n2,#0)|divide(n0,#1)|subtract(#1,const_1)|divide(n2,#3)|multiply(#2,#4)
probability
in an examination , a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 75 questions and secures 125 marks , the number of questions he attempts correctly , is :
"explanation : let the number of correct answers be x . then numbers of incorrect answers will be 75 – x we get 4 x – ( 75 – x ) × 1 = 125 on solving the equation we get x = 40 answer : b"
a ) 38 , b ) 40 , c ) 26 , d ) 25 , e ) 11
b
divide(add(125, 75), add(4, 1))
add(n2,n3)|add(n0,n1)|divide(#0,#1)|
physics
the value of x . 320 ã · 2 ã · 3 = x
320 ã · 2 ã · 3 = 320 ã — 1 / 2 ã — 1 / 3 = 160 / 3 = 53.33 correct answer : a
a ) 53.33 , b ) 46.33 , c ) 15.36 , d ) 15.45 , e ) 15.48
a
add(add(const_3, multiply(add(const_1, const_4), const_10)), divide(add(multiply(3, const_10), const_3), const_100))
add(const_1,const_4)|multiply(n2,const_10)|add(#1,const_3)|multiply(#0,const_10)|add(#3,const_3)|divide(#2,const_100)|add(#4,#5)
general
v is the volume of a cylinder ; the radius of the cylinder is 3.5 . the height of the cylinder is 650 % more than the radius . which of the following is true ?
as we see the answers are in form of range we can use approximation volume of cylinder is π r ^ 2 h given π = 22 / 7 and r = 3.5 so r ^ 2 ~ 12 and h = 6.5 * 3.5 ~ 23 so 22 / 7 * 12 * 23 ~ 868 so answer should be d . 700 < v < 900 answer : d
['a ) 100 < v < 300', 'b ) 300 < v < 500', 'c ) 500 < v < 700', 'd ) 700 < v < 900', 'e ) 900 < v < 1100']
d
multiply(multiply(const_pi, power(3.5, const_2)), multiply(3.5, divide(650, const_100)))
divide(n1,const_100)|power(n0,const_2)|multiply(#1,const_pi)|multiply(n0,#0)|multiply(#2,#3)
geometry
two employees x and y are paid a total of rs . 580 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ?
"let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 580 but x = 120 % of y = 120 y / 100 = 12 y / 10 ∴ 12 y / 10 + y = 580 ⇒ y [ 12 / 10 + 1 ] = 580 ⇒ 22 y / 10 = 580 ⇒ 22 y = 5800 ⇒ y = 5800 / 22 = rs . 263 a )"
a ) s . 263 , b ) s . 283 , c ) s . 293 , d ) s . 300 , e ) s . 383
a
divide(multiply(580, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2))
add(const_1,const_4)|multiply(#0,const_2)|add(#1,const_1)|multiply(n0,#1)|multiply(#2,const_2)|divide(#3,#4)|
general
two trains of length 120 m and 300 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ?
relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 120 + 300 = 420 m . the time required = d / s = 420 / 20 = 21 sec . answer : a
a ) 21 sec , b ) 32 sec , c ) 82 sec , d ) 20 sec , e ) 89 sec
a
divide(add(120, 300), multiply(add(42, 30), const_0_2778))
add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)
physics
two trains of length 100 m and 220 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ?
"relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 100 + 220 = 320 m . the time required = d / s = 320 / 20 = 16 sec . answer : e"
a ) 18 sec , b ) 70 sec , c ) 21 sec , d ) 20 sec , e ) 16 sec
e
divide(add(100, 220), multiply(add(42, 30), const_0_2778))
add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|
physics
what positive number , when squared , is equal to the cube of the positive square root of 18 ?
"let the positive number be x x ^ 2 = ( ( 18 ) ^ ( 1 / 2 ) ) ^ 3 = > x ^ 2 = 4 ^ 3 = 81 = > x = 9 answer e"
a ) 64 , b ) 32 , c ) 8 , d ) 4 , e ) 9
e
sqrt(power(power(18, divide(const_1, const_2)), const_3))
divide(const_1,const_2)|power(n0,#0)|power(#1,const_3)|sqrt(#2)|
geometry
a train 360 m long is running at a speed of 54 km / hr . in what time will it pass a bridge 140 m long ?
"speed = 54 * 5 / 18 = 15 m / sec total distance covered = 360 + 140 = 500 m required time = 500 * 1 / 15 = 33.33 sec answer : d"
a ) 40 sec , b ) 11 sec , c ) 88 sec , d ) 33.33 sec , e ) 10 sec
d
divide(360, multiply(subtract(54, 140), const_0_2778))
subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
210 college students were asked in a survey if they preferred windows or mac brand computers . 60 students claimed that they preferred mac to windows brand computers . one third as many of the students who preferred mac to windows , equally preferred both brands . 90 of the students had no preference . how many of the students in the survey preferred windows to mac brand computers ?
120 = 60 ( mac ) + x ( window ) + 20 ( both ) = > x = 40 answer : b
a ) 25 , b ) 40 , c ) 50 , d ) 60 , e ) 75
b
subtract(subtract(subtract(210, 90), 60), divide(subtract(subtract(210, 90), 60), const_3))
subtract(n0,n2)|subtract(#0,n1)|divide(#1,const_3)|subtract(#1,#2)
other
if a truck is traveling at a constant rate of 36 kilometers per hour , how many seconds will it take the truck to travel a distance of 600 meters ? ( 1 kilometer = 1000 meters )
"speed = 36 km / hr = > 36,000 m / hr in one minute = > 36000 / 60 = 600 meters in one sec = > 600 / 60 = 10 meters time = total distance need to be covered / avg . speed = > 600 / 10 = 60 and hence the answer : c"
a ) 18 , b ) 24 , c ) 60 , d ) 36 , e ) 48
c
multiply(divide(divide(600, 1000), 36), const_3600)
divide(n1,n3)|divide(#0,n0)|multiply(#1,const_3600)|
physics
solve this 6 + 7 = 12 8 + 9 = 16 5 + 6 = 10 7 + 8 = 14 then , 3 + 3 = ? ?
5 answer : c
a ) 11 , b ) 12 , c ) 5 , d ) 14 , e ) 18
c
add(3, subtract(3, subtract(7, subtract(12, 6))))
subtract(n2,n0)|subtract(n1,#0)|subtract(n12,#1)|add(n12,#2)
general
| x + 3 | – | 4 - x | = | 8 + x | how many solutions zwill this equation have ?
"z = | x | = x when x > = 0 ( x is either positive or 0 ) | x | = - x when x < 0 ( note here that you can put the equal to sign here as well x < = 0 because if x = 0 , | 0 | = 0 = - 0 ( all are the same ) so the ' = ' sign can be put with x > 0 or with x < 0 . we usually put it with ' x > 0 ' for consistency . a"
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
a
divide(multiply(add(4, 3), const_2), 8)
add(n0,n1)|multiply(#0,const_2)|divide(#1,n2)|
general
john and jane went out for a dinner and they ordered the same dish . both used a 10 % discount coupon . john paid a 15 % tip over the original price of the dish , while jane paid the tip over the discounted price for the coupon . if john paid $ 1.26 more than jane , what was the original price of the dish ?
the difference between the amounts john paid and jane paid is the deference between 15 % of p and 15 % of 0.9 p : 0.15 p - 0.15 * 0.9 p = 1.26 - - > 15 p - 13.5 p = 126 - - > p = 84 . answer : e .
a ) 24 , b ) 34.8 , c ) 37.8 , d ) 42 , e ) 84
e
divide(1.26, subtract(divide(15, const_100), multiply(subtract(const_1, divide(10, const_100)), divide(15, const_100))))
divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#1)|multiply(#0,#2)|subtract(#0,#3)|divide(n2,#4)
gain
if 3 / p = 8 & 3 / q = 18 then p - q = ?
"p = 3 / 8 , q = 3 / 18 = > q = 1 / 6 therefore p - q = ( 3 / 8 ) - ( 1 / 6 ) = 5 / 24 answer : a"
a ) 5 / 24 , b ) 6 / 24 , c ) 7 / 24 , d ) 8 / 24 , e ) 9 / 24
a
subtract(divide(3, 8), divide(3, 18))
divide(n0,n1)|divide(n0,n3)|subtract(#0,#1)|
general
a certain university will select 1 of 7 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 ?
"ans : 147 7 c 1 * 7 c 2 answer c )"
a ) 42 , b ) 70 , c ) 147 , d ) 165 , e ) 315
c
multiply(multiply(7, 3), 7)
multiply(n3,n5)|multiply(n1,#0)|
other
mohit sold an article for $ 18000 . had he offered a discount of 10 % on the selling price , he would have earned a profit of 8 % . what is the cost price of the article ?
"c 15000 let the cp be $ x . had he offered 10 % discount , profit = 8 % profit = 8 / 100 x and hence his sp = x + 8 / 100 x = $ 1.08 x = 18000 - 10 / 100 ( 18000 ) = 18000 - 1800 = $ 16200 = > 1.08 x = 16200 = > x = 15000"
a ) 16000 , b ) 25000 , c ) 15000 , d ) 18000 , e ) 17000
c
multiply(divide(subtract(18000, divide(multiply(18000, 10), const_100)), add(const_100, 8)), const_100)
add(n2,const_100)|multiply(n0,n1)|divide(#1,const_100)|subtract(n0,#2)|divide(#3,#0)|multiply(#4,const_100)|
gain
a is twice as fast as b . if b alone can do a piece of work in 30 days , in what time can a and b together complete the work ?
"option b explanation : a can do the work in 30 / 2 i . e . , 15 days . a and b ' s one day ' s work = 1 / 15 + 1 / 30 = ( 2 + 1 ) / 30 = 1 / 10 so a and b together can do the work in 10 days ."
a ) 13 , b ) 10 , c ) 16 , d ) 50 , e ) 23
b
inverse(add(divide(const_1, 30), multiply(divide(const_1, 30), const_2)))
divide(const_1,n0)|multiply(#0,const_2)|add(#0,#1)|inverse(#2)|
physics
a bag contains 3 blue and 5 white marbles . one by one , marbles are drawn out randomly until only two are left in the bag . what is the probability c that out of the two , one is white and one is blue ?
"the required probability c = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls . ways to select 6 out of total 8 = 8 c 6 ways to select 4 out of 5 white balls = 5 c 4 ways to select 2 out of 3 blue balls = 3 c 2 thus the required probability = ( 5 c 4 * 3 c 2 ) / 8 c 6 = 15 / 28 . d is thus the correct answer ."
a ) 15 / 56 , b ) 41 / 56 , c ) 13 / 28 , d ) 15 / 28 , e ) 5 / 14
d
divide(multiply(3, 5), divide(multiply(add(3, 5), add(5, const_2)), const_2))
add(n0,n1)|add(n1,const_2)|multiply(n0,n1)|multiply(#0,#1)|divide(#3,const_2)|divide(#2,#4)|
probability
the total cost of a vacation was divided among 4 people . if the total cost of the vacation had been divided equally among 5 people , the cost per person would have been $ 50 less . what was the total cost cost of the vacation ?
"c for cost . p price per person . c = 4 * p c = 5 * p - 250 substituting the value of p from the first equation onto the second we get p = 250 . plugging in the value of p in the first equation , we get c = 1000 . which leads us to answer choice e"
a ) $ 200 , b ) $ 300 , c ) $ 400 , d ) $ 500 , e ) $ 1000
e
multiply(multiply(5, 4), divide(50, subtract(5, 4)))
multiply(n0,n1)|subtract(n1,n0)|divide(n2,#1)|multiply(#2,#0)|
general
by mixing two brands of coffee and selling the mixture at the rate of $ 177 per kg . a shopkeeper makes a profit of 18 % . if to every 2 kg of one brand costing $ 200 per kg , 3 kg of the other brand is added , then how much per kg does the other brand cost ?
c $ 116.66 let the cost of the brand be $ x per kg . c . p . of 5 kg = ( 2 * 200 + 3 * x ) = $ ( 400 + 3 x ) s . p of 5 kg = $ ( 5 * 177 ) = $ 885 [ 885 - ( 400 + 3 x ) ] / ( 400 + 3 x ) * 100 = 18 24250 - 150 x = 3600 + 27 x 177 x = 20650 = > x = 116 2 / 3 so , cost of the other brand = $ 116.66 .
a ) $ 110.66 , b ) $ 132.66 , c ) $ 116.66 , d ) $ 126.66 , e ) $ 118.66
c
divide(subtract(multiply(add(2, 3), 177), multiply(multiply(2, 200), add(const_1, divide(18, const_100)))), multiply(3, add(const_1, divide(18, const_100))))
add(n2,n4)|divide(n1,const_100)|multiply(n2,n3)|add(#1,const_1)|multiply(n0,#0)|multiply(#3,#2)|multiply(n4,#3)|subtract(#4,#5)|divide(#7,#6)
gain
a courier charges for packages to a certain destination are 65 cents for the first 250 grams and 10 cents for each additional 100 grams or part thereof . what could be the weight in grams of a package for which the charge is $ 1.75 ?
"the charge is 65 cents for the first 250 grams . this leaves a charge of $ 1.75 - $ 0.65 = $ 1.10 the charge for the next 1000 grams is $ 1.00 which leaves a charge of $ 0.10 the weight is somewhere between 1250 and 1350 . the answer is a ."
a ) 1280 , b ) 1145 , c ) 1040 , d ) 950 , e ) 260
a
divide(add(subtract(1.75, divide(65, 100)), multiply(divide(divide(10, 100), 100), 250)), divide(divide(10, 100), 100))
divide(n2,n3)|divide(n0,n3)|divide(#0,n3)|subtract(n4,#1)|multiply(n1,#2)|add(#4,#3)|divide(#5,#2)|
general
johnny makes $ 6.75 per hour at his work . if he works 10 hours , how much money will he earn ?
6.75 * 10 = 67.50 . answer is a .
a ) $ 67.50 , b ) $ 54 , c ) $ 28.50 , d ) $ 12.50 , e ) $ 9.60
a
multiply(6.75, 10)
multiply(n0,n1)|
physics
machines x and y produce bottles at their respective constant rates . machine x produces k bottles in 6 hours and machine y produces k bottles in 3 hours . how many hours does it take machines x and y , working simultaneously , to produce 12 k bottles ?
"x rate = k / 6 y rate = k / 3 k / 6 + k / 3 = 12 k / t solving t = 24 answer e"
a ) 4 , b ) 8 , c ) 12 , d ) 18 , e ) 24
e
divide(multiply(12, divide(add(6, 3), 3)), 3)
add(n0,n1)|divide(#0,n1)|multiply(n2,#1)|divide(#2,n1)|
physics
34 , 45 , 56 , 67,78 , _ ?
answer : d
a ) 33 , b ) 82 , c ) 99 , d ) 89 , e ) 27
d
subtract(negate(67,78), multiply(subtract(45, 56), divide(subtract(45, 56), subtract(34, 45))))
negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)|
general
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 4 days . b can do the work alone in ?
"suppose a , b and c take x , x / 2 and x / 3 respectively to finish the work . then , ( 1 / x + 2 / x + 3 / x ) = 1 / 4 6 / x = 1 / 4 = > x = 24 so , b takes 12 hours to finish the work . answer : d"
a ) 19 , b ) 12 , c ) 11 , d ) 24 , e ) 114
d
multiply(add(add(const_1, const_2), const_3), 4)
add(const_1,const_2)|add(#0,const_3)|multiply(n0,#1)|
physics
a man purchased 3 blankets @ rs . 100 each , 3 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 150 . find the unknown rate of two blankets ?
"10 * 150 = 1500 3 * 100 + 3 * 150 = 750 1500 – 750 = 750 answer : a"
a ) 750 , b ) 277 , c ) 878 , d ) 450 , e ) 272
a
subtract(multiply(const_10, 150), add(multiply(3, 100), multiply(3, 150)))
multiply(n3,const_10)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(#0,#3)|
general
an amount at compound interest sums to rs . 17640 / - in 2 years and to rs . 18522 / - in 3 years at the same rate of interest . find the rate percentage ?
"the difference of two successive amounts must be the simple interest in 1 year on the lower amount of money . s . i = 18522 / - - 17640 / - = rs . 882 / - rate of interest = ( 882 / 17640 ) × ( 100 / 1 ) = > 8820 / 1764 = 5 % principal = amount / ( 1 + r / 100 ) n = 17640 / ( 1 + 5 / 100 ) 2 = 17640 / ( 21 / 20 × 21 / 20 ) = 17640 / ( 1.05 × 1.05 ) = 17640 / 1.1025 = 16000 thus the principal is rs . 16000 / - and the rate of interest is 5 % a"
a ) 5 % , b ) 6 % , c ) 7 % , d ) 8 % , e ) 9 %
a
multiply(divide(subtract(18522, 17640), 17640), const_100)
subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)|
general
how many integers from 101 to 1000 , inclusive , remains the value unchanged when the digits were reversed ?
"question is asking for palindrome first digit possibilities - 1 through 9 = 9 second digit possibilities - 0 though 9 = 10 third digit is same as first digit = > total possible number meeting the given conditions = 9 * 10 = 90 answer is e ."
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90
e
divide(1000, const_10)
divide(n1,const_10)|
general
when positive integer n is divided by 5 , the remainder is 1 . when n is divided by 7 , the remainder is 3 . what is the smallest positive integer k such that k + n is a multiple of 37 ?
"n = 5 p + 1 = 6,11 , 16,21 , 26,31 n = 7 q + 3 = 3 , 10,17 , 24,31 = > n = 37 m + 31 to get this , we need to take lcm of co - efficients of p and q and first common number in series . so we need to add 6 more to make it 37 m + 36 answer - d"
a ) 3 , b ) 4 , c ) 12 , d ) 6 , e ) 35
d
subtract(37, reminder(3, 7))
reminder(n3,n2)|subtract(n4,#0)|
general