RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: Andrew travelling to 7 cities. Gasoline prices varied from city to city. $1.75, $1.61, $1.79, $2.11, $1.96, $2.09, $1.85. What is the median gasoline price?
A.$1
B.$1.85
C.$1.92
D.$2.13
Answer: B.$1.85
Explanation: Ordering the data from least to greatest, we get:
$1.61, $1.75, $1.79, $1.85, $1.96, $2.09, $2.11
The median gasoline price is $1.85. (There were 3 states with higher gasoline prices and 3 with lower prices.)
B
Question: Simplify: 7005-7000+10
A.6505
B.6305
C.6501
D.6508
Answer: B.6305
Explanation: 7005-7000+10
=7005-(7000/10)
=7005-700
=6305.
Answer is B
Question: a bag of blue and green colored marbles has 50% green marbles, with a total of 40 marbles. how many green marbles need to be added to bring their percentage in the bag up to 80%?
A.60
B.58
C.42
D.24
Answer: A.60
Explanation: A bag of 40 marbles which is 50% green marbles means we have 20 green marbles. To get a bag that is 80% green marbles, let's use this expression: (20 + x)/(40 + x) = 0.8 with x representing the amount of green marbles that must be added to get to 80%. The expression thus represents the new number of green marbles over the total new number of marbles and this fraction should equal 80%. 60 marbles need to be added, as 80/100=0.8 or 80%
choose A
Question: A Bullock cart can travel at 10 kmph uphill road and downhill road 20 kmph. He takes 3 hours to row from lower point A to upper point B and back to A. What is the total distance traveled by bullock card in 3 hours? He must return back to base point A in remaining time.
A.80
B.20
C.30
D.40
Answer: D.40
Explanation: Upward distance traveled per hour - 10 km,
distance traveled after 2 hours = 20 km.
He remained with 1 hour travel downward, i.e. distance traveled in remaining 1 hour downward = 20 km
so total distance traveled from A to B and Back to point A = 20 km upward + 20 km downward = 40 km
(Answer : D)
Question: A and B can do a work in 12 days, B and C in 30 days and C and A in 36 days. In how many days will the work be completed, if all three of them work together?
A.180/18
B.180/15
C.180/13
D.180/12
Answer: C.180/13
Explanation: One day work of A and B = 1/12
One day work of B and C = 1/30
One day work of C and A = 1/36
2(A + B + C) = 1/12 + 1/30 + 1/36
2(A + B + C) = 13/90
(A + B + C) = 13/180
Number of days required = 180/13 days.
Answer: C
Question: A boy has 11 trousers and 11 shirts. In how many different ways can he select a trouser and a shirt?
A.121ways
B.130ways
C.90ways
D.100ways
Answer: A.121ways
Explanation: The boy can select one trouser in nine ways.
The boy can select one shirt in 12 ways.
The number of ways in which he can select one trouser and one shirt is 11 * 11
= 121 ways.
Answer: A
Question: If 3<x<6<y<9, then what is the greatest possible positive integer difference of x and y?
A.3
B.4
C.5
D.6
Answer: C.5
Explanation: 3 < x < 6 < y < 9;
3 < x
y < 9
3 + y < x + 9
y - x < 6.
Positive integer difference is 5 (for example y = 8.5 and x = 3.5)
Answer: C.
Question: If integers ending in zero are excluded, what is the unit digit of the product of the even integers from 302 to 398, inclusive?
A.2
B.4
C.6
D.8
Answer: C.6
Explanation: 302 x 304 x 306 x 308 => unit digit 4
312 x 314 x 316 x 318 => unit digit 4
392 x 394 x 396 x 398 => unit digit 4
So we have 10 numbers ending by 4 to multiply.
Given that 4 x 4 = 16, we have 5 numbers ending by 6 to multiply
Given that 6 x 6 = 36, we have 3 numbers ending by 6 to multiply
So the unit digit is 6
Answer: C
Question: The simple interest on a sum of money will be Rs.1200 after 10 years. If the principal is trebled after 5 years what will be the total interest at the end of the tenth year?
A.8981
B.3799
C.1200
D.2400
Answer: D.2400
Explanation: P --- 10 ---- 1200
P --- 5 ----- 600
3P --- 5 ----- 1800
------
=> 2400
Answer: D
Question: 4, 12, 48, 240, 1440, (...)
A.7620
B.10080
C.6200
D.10020
Answer: B.10080
Explanation: Explanation :
Go on multiplying the given numbers by 3, 4, 5, 6, 7
Answer : Option B
Question: A confectioner decides to sell all of his pastry due to the coming holiday. His pastry goods are equally divided among a group of 50 regular customers. If only 49 customers come to the bakery, each one will receive 6 less pastry goods. How much pastry does the confectioner needs to sell?
A.500.
B.412.
C.432.
D.502.
Answer: A.500.
Explanation: Pastry is divided in 50 customers equally. So,total number of pastry must be a multiple of 50
Only option A satisfies the condition, and hence is the answer
Question: For all positive integers m, [m]=3m when m is odd and [m]=(1/2)*m when m is even. What is [5]*[6] equivalent to?
A.[45]
B.[30]
C.[15]
D.[5]
Answer: C.[15]
Explanation: [5]*[6] = 15*3 = 45 = [15]
The answer is C.
Question: Eight men, ten women and six boys together can complete a piece of work in eight days. In how many days can 20 women complete the same work if 20 men can complete it in 12 days?
A.33m
B.77m
C.66m
D.22m
Answer: D.22m
Explanation: Let the number of units which can be completed by each man, each women and each boy be m/day, w/day and b/day respectively.
Work = 8(8m + 10w +6b)units = 12(20m)
10w +6b = 22m
b is unknown.
We cannot find the relation between m and w.
We cannot answer the question.
Answer: D
Question: In a village there are 4 programs organised for the festival. The festival occurs for 2 days. In how many ways that the 2 programs can be divided for 2 days?
A.4
B.10
C.6
D.5
Answer: C.6
Explanation: There are 4 programs for 2 days festival. That is Each day has 2 Programs.
The question is how many ways that the programs is chosen.
4C2=4!/2!*2!=4*3*2!/2!*2*1
4C2=6
There are 6 possible ways to organise the programs
Question: Pipes A and B fill a tank in 5 and 6 hours respectively. Pipe C empties the tank in 12 hours, if you open all the three pipes together this tank will be filled in?
A.3 9/17 hours
B.4.5 hours
C.1.25 hours
D.4 7/17 hours
Answer: A.3 9/17 hours
Explanation: Explanation:
It is given that Pipe A,B and C work simultaneously.
Pipe A and B are filling the tank and Pipe C is emptying it.
In 1 hour volume of tank filled by Pipe A: 1/5th
In 1 hour volume of tank filled by Pipe B: 1/6th
In 1 hour volume of tank emptied by Pipe C: 1/12th
Working together Pipe A B and C fill x volume of tank.
=>1/x = (1/5)+(1/6)-(1/12)
=>x=60/17 hours
ANSWER A
Question: In a class of 50 students the number of females is 2 more that 5 times the male number .How many males and females are there in the class
A.42, 8
B.50,8
C.8,42
D.8,50
Answer: A.42, 8
Explanation: Let the females be x and males be y
x+y=50--1
x=5y+2---2 This can be written as x-5y=2---3
solving 1 and 3 we get y=8
substituting y=8 in (1) we get x=42
Correct ans is A
Question: A train 240 m in length crosses a telegraph post in 16 seconds. The speed of the train is?
A.22 kmph
B.77 kmph
C.54 kmph
D.71 kmph
Answer: C.54 kmph
Explanation: S = 240/16 * 18/5 = 54 kmph
Answer:C
Question: What percent is 25 gm of 1 kg?
A.0.5 %
B.1%
C.1.5%
D.2.5%
Answer: D.2.5%
Explanation: 1 kg = 1000 gm
25/1000 × 100 = 2500/1000
=5/2 = 2.5 %
D)
Question: The price of a T.V. set worth Rs. 30000 is to be paid in 20 installments of Rs. 1000 each. If the rate of interest be 6% per annum, and the first installment be paid at the time of purchase, then the value of the last installment covering the interest as well will be?
A.22678
B.26699
C.29000
D.19000
Answer: C.29000
Explanation: Money paid in cash = Rs. 1000
Balance payment = (30000 - 1000) = Rs. 29000
Answer: C
Question: Yesterday, a pie shop sold 6 apple pies, 5 peach pies, and 6 pumpkin pies. How many pies did the shop sell in all?
A.21
B.37
C.13
D.17
Answer: D.17
Explanation: Add the number of pies
6+5+6 =17.
Answer is D.
Question: From a pack of cards two cards are drawn one after the other, with replacement. The probability that the first is a red card and the second is a king is?
A.1/26
B.1/27
C.1/29
D.1/37
Answer: A.1/26
Explanation: Let E1 be the event of drawing a red card.
Let E2 be the event of drawing a king .
P(E1 ∩ E2) = P(E1) . P(E2)
(As E1 and E2 are independent)
= 1/2 * 1/13
= 1/26
Answer: A
Question: A and B starts a business with Rs.8000 each, and after 3 months, B withdraws half of his capital . How should they share the profits at the end of the 18 months?
A.18:11
B.22:13
C.23:12
D.12:7
Answer: D.12:7
Explanation: A invests Rs.8000 for 18 months, but B invests Rs.8000 for the first 3 months and then withdraws Rs.4000. So, the investment of B for remaining 15 months is Rs.4000 only.
A : B
8000*18 : (8000*3) + (4000*15)
144000 : 84000
A:B = 12:7
ANSWER:D
Question: The ratio of number of boys and girls in a school is 2:5. If there are 210 students in the school, find the number of girls in the school?
A.150
B.250
C.300
D.370
Answer: A.150
Explanation: Let the number of boys and girls be 2x and 5x
Total students = 210
Number of girls in the school = 5*210/7 = 150
Answer is A
Question: Hector runs a one-product widget shop. Each month, his fixed costs, whether he produces anything or not, amount to x dollars. The marginal cost of producing and selling one widget is a constant y dollars, regardless of the number he produces. He generally can sell widgets for t dollars each. What is the minimum number of widgets he must produce, assuming he sells every widget he produces, to break even (neither make a profit nor incur a loss) in a given month?
A.x/(y - t)
B.x/(t - y)
C.x
D.x/t
Answer: B.x/(t - y)
Explanation: Two ways...
1) let a be the total number, so cost = a*y+x and selling price=a*t..
No profit no loss....a*y+x=a*t.....a=x/(t-y)...
B
Question: Five drainage pipes, each draining water from a pool at the same constant rate, together can drain a certain pool in 12 days. How many additional pipes, each draining water at the same constant rate, will be needed to drain the pool in 6 days?
A. 5
B. 9
C. 10
D. 12
Answer: A. 5
Explanation: This is an inverse proportional problem......
5 pipes in 12 days; So for 6 days, it will be = 12 x 5 / 6 = 10
So, 10-5 = 5
Answer A
Question: The salaries of A and B together amount to $2000. A spends 95% of his salary and B, 85% of his. If now, their savings are the same, what is A's salary?
A.$1000
B.$1250
C.$2500
D.$1500
Answer: D.$1500
Explanation: Let A's salary is x
B's salary = 2000-x
(100-95)% of x = (100-85)% of (2000-x)
x = $1500
Answer is D
Question: 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 T should the number of hours worked be increased?
A.12.5%
B.20%
C.25%
D.50%
Answer: C.25%
Explanation: 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 .8W wage and works X hours. By setting these equations equal to each other, we can determine the increase in hours worked: WH = .8WX (divide both sides by .8W) 1.25H = X We know that the new number of hours worked T will be 25% greater than the original number. The answer is C.
Question: A survey was conducted to find out how many people in a housing colony of 140 residents could swim, dance and drive a car. It was found that the number of people who could not swim was 89, the number of people who could not dance was 100 and that the number of people who could not drive a car was 91. If the number of people who could do at least two of these things, was found to be 37 and the number of people who could do all these things was found to be 6, how many people could not do any of these things?
A.43
B.44
C.45
D.46
Answer: A.43
Explanation: T=n(A)+n(B)+n(C)-n(Exactly two of the events)-2*n(All 3 Events)+n(None of the events)
T=140
n(A)=T-n(A')=140-89=51
n(B)=T-n(B')=140-100=40
n(C)=T-n(C')=140-91=49
n(Exactly two of the events)=n(At least 2 Events)-n(All 3 Events)=37-6=31
n(All 3 Events)=6
140=51+40+49 -31-2*6+n(None of the events)
n(None of the events)=140-51-40-49+31+12=43
Ans:A
Question: If m > 0, which of the following could be true?
I. m^3 > m^2
II. m^2 = m
III. m^2 > m^3
A.I only
B.III
C.IIIII
D.All of the above
Answer: D.All of the above
Explanation: 1. m^3 > m^2 Well this is true for all m > +1 (but not for values 0 < m < 1)
2. m^2 = m This is true for only one positive number, 1
3. m^2 > m^3 Again this is true only for values of m such that 0 < m < 1
So the answer is D, since there exists at least one value of m which satisfies the equation/ in equation(s)
Question: Find the odd man out. 187, 264, 336, 473, 682, 781
A.187
B.336
C.781
D.682
Answer: B.336
Explanation: In all numbers except 336, the middle digit is the sum of other two digits.Answer is B
Question: Two friends A and B apply for a job in the same company. The chances of A getting selected is 2/5 and that of B is 4/7. What is the probability that both of them get selected?
A.8/35
B.34/35
C.27/35
D.30/35
Answer: A.8/35
Explanation: EXPLANATION:
P(A) = 2/5
P(B) = 4/7
E = {A and B both get selected}
P(E) = P(A)*P(B)
= 2/5 * 4/7
= 8/35
ANSWER IS A
Question: 4 boys and 2 girls are to be seated in a row in such a way that the two girls are always together. In how many different ways can they be seated?
A.120
B.720
C.148
D.240
Answer: D.240
Explanation: Assume the 2 given students to be together (i.e one).
Now there are five students.
Possible ways of arranging them are = 5! = 120
Now, they (two girls) can arrange themselves in 2! ways.
Hence total ways = 120 × 2 = 240
Answer D
Question: If each permutation of the digits 1,2,3,4,5,6 are listed in the increasing order of magnitude, then 289thterm will be
A.361452
B.321546
C.321456
D.341256
Answer: D.341256
Explanation: Explanation :
Let's see how many numbers can be formed with the left most digit as 1
The digit '1' is placed at the 1st position (only 1 way of doing this)
Since one digit is placed at the 1st position,
any of the remaining 5 digits can be placed at 2nd position.
Since one digit is placed at the 1st position and another digit is placed
at the 2nd position, any of the remaining 4 digits can be placed at the
3rd position.
So on ...
1	5	4	3	2	1
i.e., total number of ways = (1)( 5)( 4)( 3)( 2)( 1) = 120
i.e., total count of numbers which can be formed
with the left most digit as 1 = 120
Similarly, total count of numbers which can be formed
with the left most digit as 2 = 120
Similarly, total count of numbers which can be formed
with the left most digit as 3 = 120
i.e., 240 numbers (=120 + 120) can be formed
(with left most digit as 1) or (with left most digit as 2)
Similarly, 360 numbers (=120 + 120 + 120) can be formed
(with left most digit as 1) or (with left most digit as 2)
or (with left most digit as 3)
Hence, the left most digit of the 249th number = 3
Now, let's find out how many numbers can be formed
with the left most digit as 3 and next digit as 1
The digit '3' is placed at the 1st position (only 1 way of doing this)
The digit '1' is placed at the 2nd position (only 1 way of doing this)
Any of the remaining 4 digits can be placed at 3rd position.
Since 3 digits are placed in the first three positions, any of the remaining 3 digits
can be placed at the 4th position.
Since 4 digits are placed in the first four positions, any of the remaining 2 digits
can be placed at the 5th position.
Since 5 digits are placed in the first five positions, the remaining 1 digit
can be placed at the 6th position.
1	1	4	3	2	1
i.e., total number of ways = (1)(1)(4)(3)(2)(1) = 24
i.e., Total count of numbers which can be formed
(with the left most digit as 3) and (next digit as 1) = 24
Similarly, total count of numbers which can be formed
(with the left most digit as 3) and (next digit as 2) = 24
Hence, 120 + 120 + 24 + 24 = 288 numbers can be formed
(with left most digit as 1) or (with left most digit as 2)
or (with left most digit as 3 and next digit as 1)
or (with left most digit as 3 and next digit as 2)
Hence, the 289th number is the minimum value number which is formed
with the left most digit as 3 and next digit as 4.
i.e., the number is 341256
Answer : D
Question: Siddharth wants to borrow Rs.5000 at rate of interest 6% p.a. at S.I and lend the same amount at C.I at same rate of interest for two years. What would be his income in the above transaction?
A.s.21.60
B.s.21.29
C.s.22.29
D.s.18.00
Answer: D.s.18.00
Explanation: Amount of money Siddharth borrowed at S.I at 6% p.a. for two years = Rs.5,000
He lend the same amount for C.I at 6% p.a. for two years.
=> Siddharth's income = C.I - S.I
= p[1 + r/ 100]n - p - pnr/100
= p{ [1 + r/ 100]2 - 1 - nr/100
= 5,000{ [1 + 6/100]2 - 1 - 12/100}
= 5,000 {(1.06)2- 1 - 0.12} = 5,000(1.1236 - 1 - 0.12)
= 5,000 (0.0036) = 5 * 3.6 = Rs.18.00
Answer: D
Question: If the price of a certain computer increased 30 percent from a dollars to 351 dollars, then 2a =
A.540
B.570
C.619
D.649
Answer: A.540
Explanation: Before Price increase Price = a
After 30% Price increase Price = a+(30/100)*a= 1.3a = 351 (Given)
i.e. a= 351/1.3 = $270
i.e. 2a = 2*270 = 540
Answer: option A
Question: A man has a certain number of small boxes to pack into parcles. If he packs 3, 4, 5 or 6 in a parcel, he is left with one over; if he packs 7 in a parcle, none is left over. What is the number of boxes, he may have to pack?
A.106
B.301
C.309
D.400
Answer: B.301
Explanation: Explanation:
Clearly, the required number would be such that it leaves a remainder of 1 when divided by 3, 4, 5, or 6 and no remainder when divided by 7. Thus, the number must be of the form (L.C.M of 3, 4, 5, 6) x + 1 i.e., (60x + 1 ) and a multiple of 7. Clearly, for x = 5, the number is a multiple of 7. So the number is 301.
Answer: B) 301
Question: A completes a work in 12 days and B complete the same work in 36 days. If both of them work together, then the number of days required to complete the work will be?
A.9 days
B.11 days
C.21 days
D.22 days
Answer: A.9 days
Explanation: If A can complete a work in x days and B can complete the same work in y days, then, both
of them together can complete the work in x y/ x+ y days
Therefore, here, the required number of days = 12 Ã— 36/ 48 = 9 days.
Option 'A'
Question: All the faces of cubes are painted with red colour. It is then cut into 64 equal small cubes.Find How many small cubes are there whose two adjacent faces are coloured red ?
A.0
B.8
C.16
D.24
Answer: D.24
Explanation: There are 64 small cubes,Hence one side side of the big cube is 3√64=4 cm
Number of small cubes having two adjacent faces coloured red = (x - 2) x No. of edges
= (4 - 2) x 12
= 24
Answer :D
Question: Hitesh is 40 years old and Ronnie is- 60 years old. How many years ago was the ratio of their ages 3:5 ?
A.5 years
B.10 years
C.20 years
D.15 years
Answer: B.10 years
Explanation: Suppose, the ratio was 3 : 5, x years ago.
Then,
(40-x)/(60-x)=3/5
=5 (40 - x) = 3 (60 - x)
=200-5x =180-3x
=2x = 20
= x = 10.
Answer: B
Question: Evaluate the expression
m - (m - n) : (-2) ⋅ (-5) when m = -4, n = -6
A.8
B.9
C.5
D.7
Answer: B.9
Explanation: Solution:
m - (m - n) : (-2) ⋅ (-5) = (-4) - [-4 - (-6)] : (-2) ⋅ (-5) = -4 - 2 : (-2) ⋅ (-5) = -4 - (-1) ⋅ (-5) = -4 - (+5) = -9
Answer B
Question: The average age of three boys is 15 years and their ages are in proportion 3:5:7. What is the age in years of the youngest boy?
A.8
B.9
C.7
D.5
Answer: B.9
Explanation: 3x + 5x + 7x = 45
x =3
3x = 9
Answer:B
Question: A certain board game has a row of squares numbered 1 to 100. If a game piece is placed on a random square and then moved 13 consecutive spaces in a random direction, what is the probability the piece ends no more than 13 spaces from the square numbered 49?
A.28%
B.27%
C.30%
D.37%
Answer: B.27%
Explanation: No more than 13 spaces from 49 means in the rangefrom 49-13=36 to 49+13=62, inclusive. Total numbers in this range 62-36+1=27, the probability favorable/total=27/100
Answer: B
Question: Walking at the rate of 4 kmph a man cover certain distance in 2 hr 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in.
A.12 min
B.40 min
C.60 min
D.30 min
Answer: B.40 min
Explanation: dist = 4*22/4 = 22km
new speed = 16.5 kmph
=>D/S = 11/16.5 = 40 min
ANSWER B
Question: In a friendship gang Aravind has 2 gang, in how many ways can he invite one or more of the gang to his house?
A.25
B.49
C.3
D.87
Answer: C.3
Explanation: Aravind can select one or more than one of his 8 gang.
=> Required number of ways = 2^2 – 1= 3.
C
Question: Shyam visited Ram during his brief vacation. In the mornings they both would go for yoga. In the
evenings they would play tennis. To have more fun, they indulge only in one activity per day, i.e.
either they went for yoga or played tennis each day. There were days when they were lazy and
stayed home all day long. There were 24 mornings when they did nothing, 14 evenings when they stayed at home, and a total of 22 days when they did yoga or played tennis. For how many days Shyam stayed with Ram?
A.32
B.24
C.30
D.None of these
Answer: C.30
Explanation: Explanation :
Let ,
P days : they play tennis.
Y days : they went to yoga.
T days : total duration for which Ram and Shyam were together.
=> P + Y = 22.
Also, (T - Y ) = 24 & (T - P) = 14.
Adding all of them,
2T = 22 + 24 + 14
=> T = 30 days.
Answer : C
Question: A and B began business with Rs.3000 and Rs.4000 after 8 months, A withdraws Rs.1000 and B advances Rs.1000 more. At the end of the year, their profits amounted to Rs.630 find the share of A?
A.240
B.882
C.727
D.199
Answer: A.240
Explanation: (3*8 + 2*4):(4*8 + 5*4)
8:13
8/21 * 630 = 240
Answer: A
Question: If 36 men do a work in 70 days, in how many days will 40 men do it?
A.64
B.63
C.65
D.66
Answer: B.63
Explanation: 36 * 70 = 40 * x
x = 63 days
Answer: B
Question: During a certain season, a team won 85 percent of its first 100 games and 50 percent of its remaining games. If the team won 70 percent of its games for the entire season, what was the total number of games that the team played?
A.180
B.175
C.156
D.150
Answer: B.175
Explanation: We are first given that a team won 85 percent of its first 100 games. This means the team won 0.85 x 100 = 85 games out of its first 100 games.
We are next given that the team won 50 percent of its remaining games. If we use variable T to represent the total number of games in the season, then we can say T – 100 equals the number of remaining games in the season. Thus we can say:
0.5(T – 100) = number of wins for remaining games
0.5T – 50 = number of wins for remaining games
Lastly, we are given that team won 70 percent of all games played in the season. That is, they won 0.7T games in the entire season. With this we can set up the equation:
Number of first 100 games won + Number of games won for remaining games = Total Number of games won in the entire season
85 + 0.5T – 50 = 0.7T
35 = 0.2T
350 = 2T
175 = T
Answer is B.
Question: The sum of all two digit numbers divisible by 5 is
A.945
B.545
C.745
D.645
Answer: A.945
Explanation: Required numbers are 10,15,20 . . . . . 95
This is an A.P. in which a=10,d=5 and l=95.
Let the number of terms in it be n.Then t=95
So a+(n-1)d=95.
10+(n-1)*5=95,
10+5n-5=95
5+5n=95
5n=95-5
n = 90/5
then n=18.
Required sum
=n/2(a+l)
=18/2(10+95)
=945
Answer is A.
Question: Find the principle on a certain sum of money at 5% per annum for 3 1/5 years if the amount being Rs.1160?
A.Rs.1000
B.Rs.1100
C.Rs.1010
D.Rs.10000
Answer: A.Rs.1000
Explanation: Explanation:
1160 = P [1 + (5*16/5)/100]
P= 1000
Answer: Option A
Question: A pump will fill a tank in 3 hours. Because of a leak in the tank, it took 3 hours 30 min to fill the tank. In what time the leak can drain out all the water of the tank and will make tank empty ?
A.20 hrs
B.21 hrs
C.22 hrs
D.23 hrs
Answer: B.21 hrs
Explanation: work done for 1 hr without leak = 1/3
work done with leak = 31/2 ==>7/2
work and leak in 1 hr = 2/7
work leak in 1 hr = 1/3 - 2/7 = 1/21
so 21 hrs
ANSWER B
Question: At present the ratio between the ages of Arun and Deepak is 4:3. After 6years, Arun's age will be 26 years. What is the age of Deepak at present?
A.15
B.21
C.23
D.12
Answer: A.15
Explanation: Let the present ages of Arun and Deepak be 4x years and 3x years respectively
4x+6 = 26
4x = 20
x = 5
Deepak's age = 3x = 15years
Answer is A