MU-NLPC/Calc-aqua_rat huggingface dataset:
Two dice are thrown together.What is the probability that the sum of the number on the two faces divided by 4?	Choices:
A) 3/9	B) 4/9	C) 1/9	D) 2/9	E) 5/9

Solution: n(s)=6*6= 
6*6 = 36
36   Let e be the event that the sum of the numbers on the two faces is divisible by 4   E={(1,3),(2,2),(3,1),(2,6),(3,5),(4,4),(5,3),(6,2)}   n(e)=8   p(e)=n(e)/n(s)   =8/36   =2/9   Answer is option 2/9
Answer: D
What number times (1⁄5)^2 will give the value of 5^3?  Choose the correct option:
A) 5 B) 25 C) 125 D) 625 E) 3125

Solution: x*(1/5)^2 = 5^3   x = 5^2*5^3 = 5^5 = 3125   The answer is 3125.
Answer: E
A cab on its way to Bangalore overtakes an auto also going in the same direction at 11 am. The cab reaches Bangalore at 1.30 pm and starts the return journey after taking a break of 1 hour. On its way back it meets the auto at 3 pm. At what time will the auto reach Bangalore? Choose the correct choice from the following answers:	A) 8 p.m	B) 7 p.m	C) 4 p.m	D) 1 p.m	E) 2 p.m

Solution: Explanation:   The time interval is shown below with an interval of half hour.   11 am ------- 11:30 ------- 12 ------- 12:30 ------- 1-------1:30   At 11 the cab overtakes the auto and reaches Bangalore at 1:30. Hence, the cab travels 5 segments in two and a half hour assuming the cab to be travelling at constant   speed.   On its way back, the cab commences its journey at 2:30 and meets the auto at 3 p.m. In half an hour the cab travels one segment and hence the auto in one segment away   from Bangalore.   Therefore, the auto travels 4 segments from 11 am to 3 pm assuming it to be travelling at a constant speed. Hence, the speed of auto is  4/4= 
 4/4 = 1
1 segment per hour.   As the auto is one segment away from Bangalore, it will take another 1 hour for the auto to reach Bangalore and hence the auto will reach Bangalore at 4 p.m.   ANSWER:4 p.m
Answer: C
3 men or 6 women can do a piece of work in 25 days. In how many days will 12 men and 8 women do the same work?  Choose the correct choice from the following answers:
A) 15/9 days
B) 15/4 days
C) 18/4 days
D) 11/4 days
E) 75/16 days

Solution: 3M = 6W ---- 25 days   12M + 8W -----?   24W + 8 W = 32W ---?   6W ---- 25 32 -----?   6 * 25  24  + 8= 
 24  + 8 = 32
32W ---?   6W ---- 25 32 -----?   6 * 25= 32 * x => x = 75/16 days   Answer:75/16 days
Answer: E
Two men and three women working 7 hours a day finish a work in 5 days. Four men and four women working 3 hours a day complete the work in 7 days. The number of days in which only 7 men working 4 hours a day will finish the work is?  Pick one:
A) 6 days
B) 2 days
C) 7 days
D) 5 days
E) 1 days

Solution: 5 days   2M + 3W ----- 35 h   4M + 4W ------- 21 h   7M -------? d   70M + 105W = 84M +84M   21W = 14M => 2M = 3W   4 * 35 = 7 * x => x = 20 hours   20/4= 
20/4 = 5
5 days
Answer: D
If I walk at 6 km/h, I miss the bus by 12 minutes. If I walk at 5 km/h, I reach 14 minutes before the arrival of the bus. How far I walk to reach the bus stand ?
Choose the correct choice from the following options
A) 13km
B) 12 km
C) 11 km
D) 6.5 km
E) 5.5 km

Solution: d = product of speed difference of time /difference of speed   [Here, â€“ve sign indicates before the schedule time]   d = 6Ã—5/60[12âˆ’(âˆ’14)/6-5]   â‡’ d = 13 km   13kmnswer 13km
Answer: A
The present population of a town is 60,000. The population increases annually at 10%. Find the population after 3 years.	Choose the correct choice from the following options
A) 86000  B) 79860  C) 76300  D) 82900  E) None of these

Solution: Explanation :   Solution: The required population is 60000(1 + 10/100)^3 = 60000 * 110/100 * 110/100 * 110/100 = 79,860   Answer : 79860
Answer: B
Of the 4 distinguishable wires that lead into an apartment, 2 are for cable television service, and 4 are for telephone service. Using these wires, how many distinct combinations of 3 wires are there such that at least 1 of the wires is for cable television?	Choose the correct choice from the following choices
A) 6  B) 7  C) 8  D) 9  E) 10

Solution: Given:   Total number of wires = 5   Number of cable wires = 2   Number of telephone wires = 3   To find:   Number of combination which has at least one cable wires   Solution:   No of ways of selecting'at least'1 cable wire means, we can select more than one as well. The minimum we can select is one and the maximum we can select, given the constraints that 3 wires need to be selected in total and there are 2 cable wires, is 2   Since it is a combination of wires, the arrangement is not important   6pproach 1:   Number of ways of selecting at least one cable wire in a selection of 3 wires from 5 wires =Selection 1(Number of ways of selecting one cable wire and two telephone wires )+Selection 2(Number of ways of selecting two cable wires and 1 telephone wire)   Selection 1   Number of ways of selecting one cable wire = 2C1 = 2   Number of ways of selecting 2 telephone wires = 3C2 = 3   Total = 2C1 * 3C2 = 6 ( m ways of doing something and n ways of doing something else together give m*n ways of doing - the holy grail rule in Combinatorics)   Selection 2   Number of ways of selecting one cable wire = 2C2 = 1   Number of ways of selecting 2 telephone wires = 3C1 = 3   Total = 2C2 * 3C1 = 3 ( m ways of doing something and n ways of doing something else together give m*n ways of doing - the holy grail rule in Combinatorics)   Selection 1 + Selection 2 = 9 ways of selecting 3 wires out of 5 such that at least one is a cable wire   6pproach 2   Number of ways of selecting 3 wires out of 5 such that at least one is a cable wire =Selection X(Total number of ways of selecting 3 wires from the 5) -Selection Y(total ways of selecting 3 wires such that none is a cable i.e all the three are telephone wires)   Total number of ways of selecting 3 wires out of 5 = 5C2 = 10   Number ways of selecting 3 wires such that none is a cable i.e all the three are telephone wires = 3C3 ( 3 telephone wires and we are selecting all the three at once) = 1   Selection X - Selection Y = 6   6nswer is Option 6
Answer: A
Oak trees line both sides of a street for a length of 4/10 of a kilometer. If there is 15 meters of space between the trees, and each tree is 1 meter wide, how many trees are there along the street? Choose the correct answer
A) a)	26 B) b)	23 C) c)	24 D) d)	44 E) e)	46

Solution: There is a gap of 15 m between every 2 trees , each tree is 1 m wide.   4/10 of 1 km = 400 m   Lets assume that x trees are planted So equation is   x*1 +(x-1)*15 = 400   Solve for x and we get x = 26, answer is a)	26
Answer: A
The average age of a committee of 8 members is 42 years. A member aged 55 years
retired and his place was taken by another member aged 39 years . The average age of
present committee is; Choose the correct choice from the following options:	A) 39 years	B) 38 years	C) 36 years	D) 35 years	E) 40 years

Solution: Exp. Total age of the committee = 42*8= 
42*8 = 336
336,   Total age when a member is retired   and a new one was joined= 336-55 +39, =320   Average age of present committee = 320/8= 
320/8 = 40
40.   Answer: 40 years
Answer: E
The current of a stream at 1 kmph. A motor boat goes 35 km upstream and back to the starting point in 12 hours. The speed of the motor boat in still water is?  Choose the correct option
A) 6	B) 8	C) 9	D) 4	E) 1

Solution: S = 1   M = x   DS = x + 1   US = x - 1   35/(x + 1) + 35/(x - 1) = 12   x = 6   6nswer:6
Answer: A
David mixed 10 kg of ghee at Rs. 200 per kg with 30 kg ghee at the rate of Rs. 100 per kg. At what price per kg should he sell the mixture to make a profit of 50% in the transaction? Select:  A) Rs. 180.50  B) Rs. 187.50  C) Rs. 107.50  D) Rs. 188.50  E) Rs. 187.00

Solution: Cost of Ghee = 10*200+ 30*100 /40 =5000/40= 
5000/40 = 125
125   50% of 125= 62.50   S.P. should be = 125+62.50= Rs. 187.50/ kg   Answer : Rs. 187.50
Answer: B
In the faculty of Reverse-Engineering, 226 second year students study numeric methods, 423 second year students study automatic control of airborne vehicles and 134 second year students study them both. How many students are there in the faculty if the second year students are approximately 90% of the total?  Choose the correct choice from the following answers:
A) 515.
B) 572.
C) 618.
D) 644.
E) 666.

Solution: Answer is B :572   Solution : Total number of students studying both are 423 + 226 - 134= 
423 + 226 - 134 = 515
515 (Subtracting the 134 since they were included in the both the other numbers already).   So 90% of total is 515, so 100% is approx. 572.
Answer: B
Out of first 20 natural numbers, one number is selected at random. The probability that it is either an even number or a prime number is?	Choose the correct choice from the following options
A) 17/26
B) 17/25
C) 17/22
D) 17/20
E) 17/21

Solution: n(S) = 20   n(Even no) = 10 = n(E)   n(Prime no) = 8 = n(P)   P(EᴜP) = 10/20 + 8/20 - 1/20 = 17/20.   Answer: 17/20
Answer: D
the cost if hiring a bus is shared equally by all passengers.the bus has seats for 40 passengers and the total bill comes to rs.70.37.How many seats went unoccupied?
Pick:	A) 13	B) 11	C) 10	D) 9	E) 7

Solution: 70.37/31= 
70.37/31 = 2.27
2.27   so 31 passengers give 2.27/- each..so 9 seats unoccupied   ANSWER:9
Answer: D
A Classroom contains 20 students. Eleven of the students are girls, nine are boys. If two students are randomly selected from the classroom, one after the other, what is the probability that both will be girls?
Pick: A) 1/11 B) 11/38 C) 1/3 D) 21/47 E) 8/9

Solution: Probability of getting the first girl student - 11/20   Probability of getting the second girl student - 10/19   total probability = 11/20*10/19 = 11/38   correct option 11/38
Answer: B
At 1:00 pm, a truck left city P and headed toward City Q at a constant speed of 45 km/h. One hour later, a car left City Q and headed toward City P along the same road at a constant speed of 40 km/h. If the distance between City P and City Q is 300 km, at what time will the truck and the car meet each other?	Choices
A) 4:30 B) 5:00 C) 5:30 D) 6:00 E) 6:30

Solution: At 2:00 pm, the truck and the car are 255 km apart.   The truck and the car complete a distance of 85 km each hour.   The time it takes to meet is   255/85= 
  255/85 = 3
3 hours.   They will meet at 5:00 pm.   The answer is 5:00.
Answer: B
A bag contains five white and four red balls. Two balls are picked at random from the bag. What is the probability that they both are different color?  Options
A) 5/8 B) 5/9 C) 5/2 D) 5/1 E) 5/6

Solution: Two balls can be picked from nine balls in ways.   We select one white ball and one red ball from five white balls and four red balls. This can be done ₁ ways.   The required probability = (5 * 4)/ = 20/36 = 5/9   Answer:5/9
Answer: B
Of the 100 athletes at a soccer club, 40 play defense and 64 play midfield. If at least 20 of the athletes play neither midfield nor defense, the number of athletes that play both midfield and defense could be any number between
Choices:
A) 10 to 20
B) 10 to 40
C) 24 to 40
D) 30 to 70
E) 40 to 70

Solution: First of all notice that since only 40 athletes play defense, then the number of athletes that play both midfield and defense cannot possibly be greater than 40.Eliminate D and E.   {Total} = {defense} + {midfield} - {both} + {neither}   100 = 40 + 64 - {both} + {neither}   {both} = {neither} + 4.   Since the least value of {neither} is given to be 20, then the least value of {both} is   20+4= 
  20+4 = 24
24.Eliminate A and B.   Answer: 24 to 40.
Answer: C
In a certain game, you perform three tasks. You flip a quarter, and success would be heads. You roll a single die, and success would be a six. You pick a card from a full playing-card deck, and success would be picking a diamond card. If any of these task are successful, then you win the game. What is the probability of winning?	Choices.
A) 1/48  B) 5/16  C) 11/12  D) 11/18  E) 11/16

Solution: First lets get the overview   Event 1(A) = Flipping the quarter( Probability of both winning/loosing = 1/2)   Event 2(B) = Rolling a dice( Probability of winning = 1/6 ;Probability of loosing= 1 - 1\6 =5\6)   Event 3(C) = Drawing a card( SPADES) Probability of winning = 13/52=3/4 ;Probability of loosing= 1 - 3/4 =1/4)   So now as above we have three events A,BC.   1st method(This is lengthy method)   Possible cases in which it's a win( the highlighted green event is a win and red is loose.   1.ABC= 1/2*5/6*3/4   OR   2.ABC= 1/2*1/6*3/4   OR   3.ABC= 1/2*5/6*1/4   OR   4.ABC= 1/2*1/6*3/4   OR   5.ABC= 1/2*5/6*1/4   OR   6.ABC= 1/2*5/6*1/4   OR   7.ABC= 1/2*1/6*1/4   As we now OR means add   Adding up all of them we'll get the probability of winning ie 11/16   Answer : 11/16
Answer: E
An order was placed for the supply of a carper whose length and breadth were in the ratio of 3 : 2. Subsequently, the dimensions of the carpet were altered such that its length and breadth were in the ratio 7 : 3 but were was no change in its parameter. Find the ratio of the areas of the carpets in both the cases.
Options
A) 8:6  B) 8:7  C) 8:9  D) 8:1  E) 8:2

Solution: Let the length and breadth of the carpet in the first case be 3x units and 2x units respectively.   Let the dimensions of the carpet in the second case be 7y, 3y units respectively.   From the data,.   2(3x + 2x) = 2(7y + 3y)   => 5x = 10y   => x = 2y   Required ratio of the areas of the carpet in both the cases   = 3x * 2x : 7y : 3y   = 6x2 : 21y2   = 6 * (2y)2 : 21y2   = 6 * 4y2 : 21y2   = 8:7   Answer: 8:7
Answer: B
2 students gave an examination. One of them scored 9 marks more than other and his marks were 56% of the sum of total of both their marks. What are their individual marks? Answers.
A) 42,38	B) 42,31	C) 42,32	D) 42,33	E) 42,34

Solution: Explanation:   Let marks of one of the students be X.   The marks of the other will be X+9   According to question, X+9 = 56% of X+ (X+9)   Solving, X + 9 = 56/100* (2X + 9)   => 25X + 225 = 28X + 126   => 3X = 99   => X = 33   Marks of one are 33, while of the other are 33+9= 
33+9 = 42
42.   ANSWER: 42,33
Answer: D
If the weight of 12 meters long rod is 13.4 kg. What is the weight of 6 meters long rod ?  Choose the correct choice from the following options
A) 6.7 kg. B) 10.8 kg. C) 12.4 kg. D) 18.0 kg E) None

Solution: 6.7 kg.nswer   ∵Weight of 12 m long rod = 13.4 kg   ∴ Weight of 1 m long rod = 13.4 / 12 kg   ∴ Weight of 6 m long rod = 13.4 x 6 / 12   = 6.7 kg   Option: 6.7 kg.
Answer: A
If two numbers are in the ratio 2:3. If 15 is added to both of the numbers then the ratio becomes 3:4 then find the smallest number? Choose the correct choice from the following options.
A) A)10  B) B)20  C) C)25  D) D)30  E) E)35

Solution: 2:3   2x + 15 : 3x + 15 = 3 : 4   4[2x + 15] = 3[3x + 15]   8x + 60 = 9x + 45   9x - 8x = 60 - 45= 
60 - 45 = 15
15   Then smallest number is= 2   2x = 30   Correct Option 30
Answer: D
A 25 cm wide path is to be made around a circular garden having a diameter of 4 meters. Approximate area of the path is square meters is?  Pick one:
A) 3.34 sq m	B) 4.98 sq m	C) 2.88 sq m	D) 2.98 sq m	E) 2.56 sq m

Solution: 3.34 sq mrea of the path = 3.34 sq mrea of the outer circle - 3.34 sq mrea of the inner circle   = ∏{4/2 + 25/100}2 - ∏[4/2]2   = ∏[2.252 - 22] = ∏(0.25)(4.25) { (a2 - b2 = (a - b)(a + b) }   = (3.14)(1/4)(17/4)   = 53.38/16   = 3.34 sq m   3.34 sq mnswer:3.34 sq m
Answer: A
If a trader sold two cars each at Rs. 325475 and gains 11% on the first and loses 11% on the second, then his profit or loss percent on the whole is?	Pick one:  A) 1.21%  B) 1.74%  C) 1.84%  D) 1.47%  E) 1.24%

Solution: SP of each car is Rs. 325475, he gains 11% on first car and losses 11% on second car.   In this case, there will be loss and percentage of loss is given by   = [(profit%)(loss%)]/100   = (11)(11)/100 % = 1.21%   1.21%nswer:1.21%
Answer: A
If a and b are both odd prime numbers and a < b, then how many different positive integer factors does 2ab have?	Pick one:
A) 3 B) 4 C) 6 D) 8 E) 12

Solution: a and b are both odd prime numbers- It means either a or b is not 2   and since prime numbers have only two factors - 1 and the number itself   a and b each will have (1+1)= 
(1+1) = 2
2 factors   Hence 2ab will have (1+1)(1+1)(1+1)= 8 factors   8 is the answer
Answer: D
A tank is filled to one quarter of its capacity with a mixture consisting of water and sodium chloride. The proportion of sodium chloride in the tank is 30% by volume and the capacity of the tank is 24 gallons. If the water evaporates from the tank at the rate of 0.4 gallons per hour, and the amount of sodium chloride stays the same, what will be the concentration of water in the mixture in 6 hours? Choose the correct answer:
A) 30% B) 35% C) 40% D) 45% E) 50%

Solution: The number of gallons in the tank is (1/4)24 = 6 gallons   The amount of sodium chloride is 0.3(6) = 1.8 gallons   At the start, the amount of water is 0.7(6) = 4.2 gallons   After 6 hours, the amount of water is 4.2 - 0.4(6) = 1.8 gallons   The concentration of water is 1.8 / (1.8+1.8) = 1.8/3.6 = 50%   The answer is 50%.
Answer: E
A corporation triples its annual bonus to 50 of its employees. What percent R of the employees’ new bonus is the increase?
Select
A) 50%	B) 662⁄3%	C) 100%	D) 200%	E) 300%

Solution: R of the employees’ new bonus is the increase Hence 662⁄3%.
Answer: B
The sale price sarees listed for Rs.400 after successive discount is 10% and 5% is?
Choose the most appropriate option
A) 348 B) 177 C) 342 D) 165 E) 189

Solution: 400*(90/100)*(95/100)= 
400*(90/100)*(95/100) = 342
342   Answer:342
Answer: C
A tradesman by means of his false balance defrauds to the extent of 28%? in buying goods as well as by selling the goods. What percent does he gain on his outlay?	Choose the correct answer:
A) 37
B) 27
C) 68.84
D) 78.45
E) 62.89

Solution: Explanation:   g% = 28 + 28 + (28*28)/100= 
28 + 28 + (28*28)/100 = 1_596/25 = around 63.84
63.84%   Answer: 68.84
Answer: C
There are 5person among whom 2are brother. The total no.of ways in which these persons can be seated around a round table so that exactly 1person sit between the brothers, is equal to?  Options:  A) 4!*2!  B) 2!*2!  C) 6!*1!  D) 2!*1!  E) 5!*1!

Solution: Total number of ways = 2!*2!.   2!*2!
Answer: B
A lemonade stand sold only small and large cups of lemonade on Tuesday. 1/5 of the cups sold were small and the rest were large. If the large cups were sold for 7/6 as much as the small cups, what fraction of Tuesday's total revenue was from the sale of large cups?  Choices:  A) 7/16  B) 7/15  C) 10/21  D) 14/17  E) 1/2

Solution: Let the total # of cups be 10.   # of small cups sold 1/5*10= 
1/5*10 = 2
2;   # of large cups sold  10-2= 
 10-2 = 8
8;   Let the price of small cup be $6, then the price of larges cup would be 7/6*6= 
7/6*6 = 7
$7;   Revenue from small cups: 2*$6=$12;   Revenue from large cups cups: 8*$7  2* 6= 
  2* 6 = 12
$12;   Revenue from large cups cups: 8*$7=$56;   Fraction of total revenue from large cups: 56/(56+12) 8* 7= 
 8* 7 = 56
$56;   Fraction of total revenue from large cups: 56/(56+12)=14/17.   Answer: 14/17.
Answer: D
If a and b are the roots of the equation x2 - 8x + 8 = 0, then the value of a2 + b2 is:  Choose the correct choice from the following answers: A) 36 B) 48 C) 17 D) 6 E) 5

Solution: Sol.(b) The sum of roots = a+ b = 8   Product of roots = ab = 8   Now, a2 + b2 = (a + b)2 - 2ab = 64 - 16= 
64 - 16 = 48
48   Answer 48
Answer: B
Amit and Ian paint a wall in alternating shifts. First Amit paints alone, then Ian paints alone, then Amit paints alone, etc. During each of his shifts, Amit paints 1/2 of the remaining unpainted area of the wall, while Ian paints 1/3 of the remaining unpainted area of the wall during each of his shifts. If Amit goes first, what fraction of the wall's area will remain unpainted after Amit has completed his 7th shift?	Pick:
A) 1/27  B) 1/54  C) 1/81  D) 1/162  E) 1/1458

Solution: Fraction of area unpainted after FIRST shift of Amit = 1-(1/2) = 1/2   Fraction of area unpainted after FIRST shift of Ian = (1/2)-(1/3)(1/2) = (2/3)(1/2)   Fraction of area unpainted after S1/1458COND shift of Amit = (1/2)(2/3)(1/2)   Fraction of area unpainted after S1/1458COND shift of Ian = (2/3)(1/2)(2/3)(1/2)   Fraction of area unpainted after THIRD shift of Amit = (1/2)(2/3)(1/2)(2/3)(1/2)   Fraction of area unpainted after THIRD shift of Ian = (2/3)(1/2)(2/3)(1/2)(2/3)(1/2)   Fraction of area unpainted after FORTH shift of Amit = (1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)   Fraction of area unpainted after FORTH shift of Ian = (2/3)(1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)   Fraction of area unpainted after FIFTH shift of Amit = (1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)   Fraction of area unpainted after FIFTH shift of Ian = (2/3)(1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)   Fraction of area unpainted after SIXTH shift of Amit = (1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)   Fraction of area unpainted after SIXTH shift of Ian = (2/3)(1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)   Fraction of area unpainted after S1/1458V1/1458NTH shift of Amit = (1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)(2/3)(1/2)=1458   Answer: option 1/1458
Answer: E
A train running at a speed of 60 kmph crosses a pole in 9 seconds.What is the length of the train?	Options:
A) 120 m
B) 180 m
C) 190 m
D) 150 m
E) 160 m

Solution: 60 kmph=50/3 m/sec   50/3*9= 
50/3*9 = 150
150 m   ANSWER:150 m
Answer: D
In a group of hats consisting of only blue hats, green hats, and purple hats, the ratio of blue hats to green hats to purple hats is 7:4:12. If there are a total of 138 hats in this group, how many of these hats are not blue? Select:
A) 96
B) 42
C) 48
D) 64
E) 76

Solution: B:G:P = 7:4:12 --> total of 23 parts.   Not blue = (G + P)/(total)*92 = 16/23*138= 
16/23*138 = 96
96.   96nswer: 96.
Answer: A
123 school girls are to be transported in small vans. Each van can carry 8 girls only. What is the smallest possible number of vans that are needed to transport all 123 school girls?  Pick one
A) 4	B) 8	C) 16	D) 24	E) 32

Solution: To find the number of vans, we divide 123 by 8.   123 *8 = 15 and remainder = 3   So 15 vans are needed to transport 15 ? 8 = 120 girls, and 1 van is needed to transport the 3 remaining girls. A total of 16 vans are needed.   correct answer 16
Answer: C
A and B together can do a work in 6 days. If A alone can do it in 15 days. In how many days can B alone do it?  Choose the most appropriate option.
A) 11	B) 10	C) 21	D) 20	E) 25

Solution: 10   10   1/6 – 1/15 = 1/10 => 10
Answer: B
A boat can move upstream at 22 kmph and downstream at 42 kmph, then the speed of the current is?
Choose the correct choice from the following choices:  A) 5  B) 6  C) 7  D) 10  E) 3

Solution: US = 22   10S = 42   M = (42 - 22)/2= 
(42 - 22)/2 = 10
10.Answer:10
Answer: D
A sum of money at simple interest amounts to Rs. 2240 in 2 years and to Rs. 2600 in 5 years. What is the principal amount Choose the correct choice from the following answers
A) 1000	B) 1500	C) 2000	D) 2500	E) None of these

Solution: Explanation:   SI for 3 year = 2600-2240= 
2600-2240 = 360
360   SI for 2 year 360/3 * 2= 
360/3 * 2 = 240
240   principal= 2240 - 240= 
2240 - 240 = 2_000
2000   Option 2000
Answer: C
If M and N are non-zero integers, and M/N > 1, then which of the following must be true?	Choose the most appropriate option.
A) M > N
B) 2M > N
C) MN > N
D) M^2< N^2
E) M^3 < N^3

Solution: MN > N
Answer: C
A man saves 20% of his monthly salary. If an account of dearness of things he is to increase his monthly expenses by 10%, he is only able to save Rs. 500 per month. What is his monthly salary?
Choices
A) Rs.4500	B) Rs.4000	C) Rs.4167	D) Rs.4200	E) Rs.3000

Solution: Income = Rs. 100   Expenditure = Rs. 80   Savings = Rs. 20   Present Expenditure 80+80*(10/100) = Rs.88   Present Savings = 100 – 88 =Rs. 12   If savings is Rs. 12, salary = Rs. 100   If savings is Rs.500, salary = 100/12*500 = 4167   Answer : Rs.4167
Answer: C
What should be added to 4750 so that it may become a perfect square?	Pick one:
A) 11
B) 15
C) 16
D) 17
E) 18

Solution: 69x69=4761   4761-4750= 
4761-4750 = 11
11   If added to 11 get perfect square   answer=11
Answer: A