RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: Rahim travels 4 Kms in 15 minutes and another 4 Kms in 5 mins. Deepak travels at the average speed of Rahim. Then what could be the distance covered by him in 50 minutes:
A.10.m.
B.12 km.
C.20 km.
D.13.5km.
Answer: C.20 km.
Explanation: Distence=8kms,time=20mins;
the ditence covered by him in 50mins=50*8/20=20km
ANSWER:C
Question: If the positive integer x is a multiple of 3 and the positive integer y is a multiple of 3, then xy must be a multiple of which of the following?
I. 3
II. 12
III. 9
A.II only
B.I and III only
C.I and II only
D.II and III only
Answer: B.I and III only
Explanation: x=3a
y=3b
xy=9ab
where a,b are integers
Hence , xy will always be divisible by 3,9 .
B is the answer.
Question: A man buys a cycle for Rs. 800 and sells it at a loss of 15%. What is the selling price of the cycle?
A.s. 1090
B.s. 1160
C.s. 680
D.s. 520
Answer: C.s. 680
Explanation: S.P. = 85% of Rs. 800 = Rs.85/100x 800 = Rs. 680
ANSWER :C
Question: If n divided by 7 has a remainder of 2, what is the remainder when 5 times n is divided by 7?
A.1
B.2
C.3
D.5
Answer: C.3
Explanation: As per question => N=7P+2 for some integer p
hence 5N => 35Q + 10 => remainder =>10 for some integer Q
Alternatively => N=2 > 5N=> 10 => 10 divided by 7 will leave a remainder 3
hence C
Question: If the simple interest on a sum of money for 2 years at 5% per annum is Rs.50, what is the compound interest on the same at the same rate and for the same time?
A.Rs.51.25
B.Rs.52
C.Rs.54.25
D.Rs.60
Answer: A.Rs.51.25
Explanation: Solution
Sum	=Rs.(50 x 100/2x5)
= Rs. 500.
Amount	=[Rs.(500 x(1+5/100)²]
= Rs(500x21/20x21/20).
=Rs. 551.25
C.I	= Rs. (551.25 - 500)
= Rs. 51.25
Answer A
Question: In how much time will a train of length 130 m, moving at 36 kmph cross an electric pole?
A.sec
B.sec
C.sec
D.sec
Answer: D.sec
Explanation: Convert kmph to mps. 36 kmph = 36 * 5/18 = 10 mps.
The distance to be covered is equal to the length of the train.
Required time t = d/s = 130/10 = 13 sec.
Answer:D
Question: The original price of a certain TV set is discounted by s percent, and the reduced price is then discounted by 2s percent. If P is the original price of the TV Set, which of the following represents the price of the television set after the two successive discounts?
A.P(1 - 0.03x + 0.02x^2)
B.P(1 - 0.03s + 0.0002s^2)
C.P(1 - 0.03x + 0.002x^2)
D.P(1 - 2x^2)
Answer: B.P(1 - 0.03s + 0.0002s^2)
Explanation: Youcansolve this problem with number plugging.
Say the original price was $10 and s=50. Then after the first reduction the price would become $5 and after the second reduction of 2*50=100% the rprice would become $0.
Now, since P is not zero, then the expression in the brackets must be zero for s=50. Only answer choice B works.
Answer: B.
On this problem - number pluggin is not giving me the answer.. I initially used s = 10, then 2s = 20 and P = 100. Answer should after both consecutive discounts = 72. I plug in the respective values and I keep getting 68. Can you double check my math.
100 (1-0.03(10) + 0.0002 (10)^2)
100 (1-0.3 + 0.0002 (100))
100 (0.7 + 0.02)
100 (0.68) = 68????
double check my math. Am I missing something? I also plugged in your numbers and still did not get zero as final answer with choice B..
Question: A, B and C play a cricket match. The ratio of the runs scored by them in the match is A:B = 1:3 and B:C = 1:5. If the total runs scored by all of them are 95, the runs scored by C are?
A.20.23
B.20.13
C.75
D.20.93
Answer: C.75
Explanation: A:B = 1:3
B:C = 1:5
A:B:C = 1:3:15
15/19 * 95 = 75
ANSWER:C
Question: There are 15 stamps from which two stamps will be selected. Tom has five stamps which are same as five of the ten stamps to be selected. What is the probability that two stamps selected will not be one of the stamps that Tom has?
A.3/7
B.8/15
C.6/17
D.5/18
Answer: A.3/7
Explanation: the highlighted portion is NOT clear and poorly written, But I believe you mean that Tom has five stamps of the 15 from which two are being selected..
so we have ONLY 10 which can be selected..
P=10/15∗9/14=3/7
ANSWER:A
Question: During the first two weeks of June, the total rainfall in Springdale was 40 inches. If the rainfall during the second week was 1.5 times the rainfall during the first week, what was the rainfall in inches during the second week of June?
A.15
B.18
C.21
D.24
Answer: D.24
Explanation: Let x be the rainfall in the first week. Then 1.5x was the rainfall in the second week.
2.5x = 40
x = 16
The rainfall during the second week was 1.5 * 16 = 24 inches
The answer is D.
Question: On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:
A.Rs. 45
B.Rs. 50
C.Rs. 55
D.Rs. 60
Answer: D.Rs. 60
Explanation: (C.P. of 17 balls) - (S.P. of 17 balls) = (C.P. of 5 balls)
C.P. of 12 balls = S.P. of 17 balls = Rs.720.
C.P. of 1 ball = Rs.720/12	= Rs. 60.
Answer: Option D
Question: A sock drawer contains 3 socks, each one a different solid color. The colors of the socks are red, orange and violet. If a person reaches into the drawer and pulls out two socks, how many different color combinations are possible in the selected pair of socks?
A.12
B.15
C.21
D.3
Answer: D.3
Explanation: Since the order in which we select the socks does not matter , we can use combinations.
We can select 2 socks from 3 socks in 3C2 ways.
3C2 = 3
Answer : D
Question: A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 265 m long?
A.50
B.99
C.88
D.77
Answer: A.50
Explanation: :
Speed = 45 * 5/18 = 25/2 m/sec
Total distance covered = 360 + 265 = 625 m
Required time = 625 * 2/25 = 50 sec
Answer:A
Question: -20 , -16 , -12 , -8 ....
In the sequence above, each term after the first is 3 greater than the preceding term. Which of the following could not be a term in the sequence?
A.0
B.200
C.441
D.663
Answer: B.200
Explanation: Ans : B
Since 762 is not divisible by 3 , all the other options are divisible by 3
Question: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value T of the product xy?
A.	-12
B.	-3
C.	0
D.	2
Answer: C.	0
Explanation: How to deal with inequalities involving absolute values? First example shows us the so callednumber case
In this case we have |y + 3| ≤ 3 which is generalized |something| ≤ some number. First we solve as if there were no absolute value brackets:
y + 3 ≤ 3
y ≤ 0
So y is 0 or negative
Second scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality,ANDflip the sign:
y + 3 >= -3
y >= -6
Therefore we have a possible range for y: -6=<y<=0
Ok, so far so good, we're half way through. What about x?
Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x.
You can say that 2y + 6 is a multiple of 3 (=3x). So all values which must be integer must also satisfy this constraint. I'm just saying that, so it's easier to evaluate all the possible numbers (-6, -3, 0). If you plug in y = 0, x will be 2 and xy = 0 as the lowest possible value T.
Hence, Answer Choice C is the one to go.
Question: In a certain alphabet, 8 letters contain a dot and a straight line. 24 letters contain a straight line but do not contain a dot. If that alphabet has 40 letters, all of which contain either a dot or a straight line or both, how many letters contain a dot but do not contain a straight line?
A.4
B.8
C.14
D.20
Answer: B.8
Explanation: We are told that all of the letters contain either a dot or a straight line or both, which implies that there are no letters without a dot and a line (no line/no dot box = 0).
First we find the total # of letters with lines: 8 +24 = 32;
Next, we find the total # of letters without line: 40 - 32= 8;
Finally, we find the # of letters that contain a dot but do not contain a straight line:8- 0 = 8.
B
Question: A man invests Rs. 6,000 at the rate of 5% per annum. How much more should he invest at the rate of 8%, so that he can earn a total of 6% per annum?
A.Rs. 1200
B.Rs. 1300
C.Rs. 1500
D.Rs. 3000
Answer: D.Rs. 3000
Explanation: Explanation :
Interest on Rs.6000 at 5% per annum = ( 6000 × 5 × 1) / 100 = Rs. 300
Let his additional investment at 8% = x
Interest on Rs.x at 8% per annum = ( x × 8 × 1 ) / 100 = 2x/25.
To earn 6% per annum for the total, interest = (6000 + x) × 6 × 1/100.
=> 300 + 2x/25 = (6000 + x) × 6 × 1/100.
=> 30000 + 8x = (6000 + x) × 6.
=> 30000 + 8x = 36000 + 6x.
=> 2x=6000.
=> x=3000.
Answer : D
Question: When a certain number X is divided by 63, the remainder is 11. What is the remainder when X is divided by 9?
A.2
B.3
C.4
D.5
Answer: A.2
Explanation: When a certain number X is divided by 63, the remainder is 11. What is the remainder when X is divided by 9?
Putting a value say x = 11 we get remainder as 11 when divided by 63.
When 11 divided by 9 we get 2 as remainder.
A is the answer.
Question: Three pipes of same capacity can fill a tank in 8 hours. If there are only two pipes of same capacity, the tank can be filled in.
A.17 hours
B.12 hours
C.16 hours
D.24 hours
Answer: B.12 hours
Explanation: Explanation:
The part of the tank filled by three pipes in one hour = 1/8
=> The part of the tank filled by two pipes in 1 hour = 2/3 * 1/8 = 1/12.
The tank can be filled in 12 hours.
ANSWER IS B
Question: Three unbiased coins are tossed. What is the probability of getting at most two heads?
A.3/4
B.1/4
C.3/8
D.7/8
Answer: D.7/8
Explanation: Explanation:
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
P(E) =n(E)/n(S)= 7/8	.
ANSWER : D
Question: Two numbers are in the ratio 3:5. If 9 be subtracted from each, they are in the ratio of 9:17. The first number is:
A.36
B.77
C.88
D.55
Answer: A.36
Explanation: (3x-9):(5x-9) = 9:17
x = 12 => 3x = 36
Answer: A
Question: How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2?
A.2,000
B.1,900
C.1,800
D.1,600
Answer: C.1,800
Explanation: Digit A can be filled in 9 ways (excluding 0)
Digit B can be filled in 10 ways
Digit C can be filled in 10 ways
Digit D can be filled in 2 ways (because of modulus A-D )
The answer will be 9*10*10*2 = 1800
Ans C
Question: The time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream. If the speed of the boat in still water is 45 kmph, find the speed of the stream?
A.15
B.88
C.14
D.12
Answer: A.15
Explanation: The ratio of the times taken is 2:1.
The ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(2-1) = 3/1 = 3:1
Speed of the stream = 45/3 = 15 kmph.
Answer:A
Question: Village X has a population of 68000, which is decreasing at the rate of 1200 per year. Village Y has a population of 42000, which is increasing at the rate of 800 per year .in how many years will the population of the two villages be equal?
A.13
B.15
C.16
D.12
Answer: A.13
Explanation: Let the population of two villages be equal after p years
Then,68000-1200p=42000+800p
2000p=26000
p=13
Answer is A.
Question: Shweta buys a product at 25% discount rate. At what % over the cost price should she sell the product to make an overall 25% profit over the listed price?
A.66.67%
B.66.69%
C.66.63%
D.66.64%
Answer: A.66.67%
Explanation: Let the listed price of the product be x
Hence, at 25% discount rate the Cost Price will become: 3x/4
Now to make an overall profit of 25% over listed price:
SP = CP+(25/100)CP
=> SP = x+x/4
=>SP = 5x/4
=>Selling Price over Cost price: (5x/4)-(3x/4)
=>x/2
=>%Selling price over Cost price: (x/2)/(3x/4)*100
=>
ANSWER: A
Question: The ratio of the present ages of P and Q is 3:4. 5 years ago, the ratio of their ages was 5:7. Find the their present ages?
A.30,40
B.25,30
C.50,60
D.20,40
Answer: A.30,40
Explanation: Answer: Option A
Their present ages be 3X and 4X.
5 years age, the ratio of their ages was 5:7, then (3X - 5):(4X - 5) = 5:7
X = 35 - 25 => X = 10.
Their present ages are: 30, 40.
Question: The number of cars present at a particular time was measured at 2,999 different locations on Tuesday and on Wednesday. The number of locations that had more cars on Wednesday was 10% higher than the number of locations that had more cars on Tuesday. How many of the locations had more cars on Tuesday?
A.1,428
B.1,633
C.1,860
D.2,460
Answer: A.1,428
Explanation: The question states that all these 2,999 locations were measured twice, once on each day.
It also breaks the 2,999 into two groups, the higher-on-Tuesday locations and the higher-on-Wednesday locations, which we can call T and W.
Therefore, we can write two equations:
For each of the first and second day count of locations, T + W = 2999
But, for the second day count, W = 1.10 T
i.e. for Second day count, T + 1.10T = 2999
i.e. 2.10 T = 2999
i.e. T = 1428
A
Question: On dividing a certain number by 234, we get 43 as remainder. If the same number is divided by 13, what will be the remainder?
A.6
B.9
C.4
D.7
Answer: C.4
Explanation: Description :
Solution : suppose that on dividing the given number by 234,
we get quotient=x and remainder= 43
then, number= 234*x+43----> (1).
=> (13*18x)+(13*3)+4
=> 13*(18x+3)+4.
So, the number when divided by 13 gives remainder=4.
Answer :(C)
Question: The average mark of the students of a class in a particular exam is 80. If 5 students whose average mark in that exam is 50 are excluded, the average mark of the remaining will be 90. Find the number of students who wrote the exam.
A.20
B.27
C.25
D.99
Answer: A.20
Explanation: Let the number of students who wrote the exam be x.
Total marks of students = 80 x.
Total marks of (x - 5) students = 90(x - 5)
80x - (5 * 50) = 90(x - 5)
200 = 10x => x = 20.Answer: A
Question: Two pipes A and B can fill a tank in 4 and 5 hours respectively. If they are turned up alternately for one hour each, the time taken to fill the tank is?
A.4 hrs 27 min
B.4 hrs 24 min
C.4 hrs 74 min
D.4 hrs 54 min
Answer: B.4 hrs 24 min
Explanation: 1/4 + 1/5 = 9/20
20/9 = 2 2/9
9/20 * 2 = 9/10 ---- 4 hours
WR = 1 - 9/10 = 1/10
1 h ---- 1/4
? ----- 1/10
2/5 * 60 = 24 = 4 hrs 24 min
Answer:B
Question: The perimeters of two squares are 80 cm and 64 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares.
A.42cm
B.44cm
C.46cm
D.48cm
Answer: D.48cm
Explanation: Explanation:
Side of first square = (80/4) = 20 cm;
Side of second square = (64/4)cm = 16 cm.
Area of third square = [(20)2 - (16)2] cm2 = (400 - 256) cm2 = 144 cm2.
Side of third square = (144)(1/2) cm = 12 cm.
Required perimeter = (12 x 4) cm = 48 cm.
Answer: Option D
Question: The work done by a woman in 8 hours is equal to the work done by a man in 6 hours and by a boy in 12 hours. If working 6 hours per day, 9 men can complete a work in 6 days then in how many days can 12 men, 12 women and 12 boys together finish the same work by working 8 hours per day?
A.1 1⁄3 days
B.3 2⁄3 days
C.3 days
D.1 1⁄2 days
Answer: D.1 1⁄2 days
Explanation: 8W = 6M = 12B
12M + 12W + 12B Þ 12M + 9M + 6M = 27M
\9 men can complete the work by working 1 hour per day in 6 × 6 days
\27 men working 8 hours per day = 6×6×9/27×8=112days
Answer D
Question: A, B and C started a business with investment of Rs. 8000. Rs. 20000 and Rs. 32000. What is the share of B in the total profit of Rs. 5400 at the end of 1 year?
A.Rs. 2880
B.Rs. 720
C.Rs. 1800
D.Rs. 1650
Answer: C.Rs. 1800
Explanation: Explanation:
Ratio of capitals of A, B and C
= 8000 : 20000 : 32000 = 2 : 5 : 8
.: B's share = Rs. [(5/15) × 5400] = Rs. 1800
Answer: Option C
Question: Integer m has 4 different prime factors and n has 3 different prime factors. If m and n has the greatest common factor of 15, how many different prime factors does mn have?
A. 4
B. 5
C. 6
D. 7
Answer: B. 5
Explanation: As m and n has GCF of 15, 3*5 should be common factor for both m and n.
m=3*5*p1*p2
n=3*5*p3
mn=3^2*5^2*p1*p2*p3
mn has prime numbers of 3,5,p1,p2p3
B
Question: A, Band C can do a piece of work in 11 days, 5 days and 55 days respectively, working alone. How soon can the work be done if A is assisted by Band C on alternate days?
A.5 days
B.8 days
C.9 days
D.10 days
Answer: A.5 days
Explanation: (A+B)'s 1 day's work = 1/11+1/5 = 16/55
(A+C)'s 1 day's work = 1/11+1/55 = 6/55
work done in 2 day's = 16/55+6/55 = 2/5
2/5th work done in 2 days
work done= 5/2*2 = 5 days
ANSWER:A
Question: Which one of the given numbers is divisible by 11?
A.4823718
B.8423718
C.8432718
D.4132788
Answer: D.4132788
Explanation: Explanation:
4132788 = (4+3+7+8) – (1+2+8) = 11
So, the given number is divisible by 11
Answer:D
Question: Cn = 2Cn-1 + 4 and Qn = 4Qn-1 + 8 for all n > 1. If C5 = Q4 and C7 = 316, what is the first value of n for which Qn is an integer?
A. 1
B. 2
C. 3
D. 4
Answer: C. 3
Explanation: Assuming I have understood the symbols used correctly answer is C- 3
C7 = 2S6 +4 = 316
C6 = 312/2 = 156 = 2C5 + 4
C5 = 152/2 = 76 = Q4
Q4 = 4Q3 + 8 = 76
Q3 = 68/4 = 17
Q3 = 4Q2 + 8 = 17
Q2 = 9/4 = 2.25
Q2 = 4Q1 + 8
Q1 will not be integer
Question: RAJA complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
A.200 km
B.250 km
C.224 km
D.255 km
Answer: C.224 km
Explanation: consider x -- > (1/2)x/21 + (1/2)x/24 = 10
==> x/21+x/24 = 20
15x = 168 * 20
==> 224km
ANSWER C
Question: Find avrg speed if a man travels at speed of 18km/hr up and 24km/hr dawn at an altitude of 300m.
A.25.8
B.26.8
C.27.8
D.20.5
Answer: D.20.5
Explanation: avg speed=2*x*y/(x+y)
=2*18*24/(18+24)=20.5
ANSWER:D
Question: A train 800 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is?
A.300
B.400
C.500
D.600
Answer: C.500
Explanation: Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (800 + x)/60 = 65/3
x = 500 m.
Answer: Option C
Question: If the sides of a triangle are given as 5*x, 5*(x+1), 5*(x+2), then what could be the possible value of the product of their sides.
A.1000
B.2000
C.3000
D.4000
Answer: C.3000
Explanation: Since the sides are consecutive multiples of 5, the product will be a multiple of 125*2*3(=750).
750*4=3000
Answer:C
Question: Convert 350 inches into centimeter?
A.812cm
B.820cm
C.850cm
D.889cm
Answer: D.889cm
Explanation: 1 inch = 2.54cm
350 inches = 100*2.54 = 889cm
Answer is D
Question: A certain box has 13 cards and each card has one of the integers from 1 to 13 inclusive. Each card has a different number. If 2 different cards are selected at random, what is the probability that the sum of the numbers written on the 2 cards is less than the average (arithmetic mean) of all the numbers written on the 13 cards?
A.1/7
B.1/13
C.1/15
D.1/22
Answer: B.1/13
Explanation: The average of the numbers is 7
The total number of ways to choose 2 cards from 13 cards is 13C2 = 78.
The ways to choose 2 cards with a sum less than the average are:
{1,2},{1,3},{1,4},{1,5},{2,3},{2,4}
The probability is 6/78 = 1/13
The answer is B.
Question: If a clock takes 7seconds to strike 7, how long will the same clock take to strike 10?
A.10.5
B.11.5
C.12.5
D.13.5
Answer: A.10.5
Explanation: If a clock takes 7seconds to strike 7 , there are 6 time intervals of duration 7/6 secs.
At 10, there are 9 time intervals of 7/6 secs having total duration of 9*7/6= 10.5 secs for 10 strikes
ANSWER:A
Question: If one cat eats 1.5 rats in 1.5 days, then in howmany days 3 cats will eat 3 rats??
A.1 Day
B.2 Days
C.3 Days
D.4 Days
Answer: A.1 Day
Explanation: 1Cat ------ 3/2 Rat --------- 3/2 Days
1 Cat ------(3/2)*2 Rat -----(3/2)*2 Days (Rat doubled, days doubled)
1*3 Cat----- 3 Rat -----------3/3=1 Day (Cat*3 , Same rat , Days/3 )
ANSWER:A
Question: A contractor undertakes to built a walls in 50 days. He employs 20 peoples for the same. However after 25 days he finds that only 40% of the work is complete. How many more man need to be employed to complete the work in time?
A.25
B.30
C.35
D.30
Answer: D.30
Explanation: 20 men complete 0.4 work in 25 days.
Applying the work rule, m1 × d1 × w2 = m2 × d2 × w1
we have,
20 × 25 × 0.6 = m2 × 25 × 0.4
or m2 = 20×25×0.6/25×0.4 = 30 men
AnswerD
Question: A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are drawn at random, what is the probability that at least one is green?
A.23/35
B.23/38
C.23/18
D.53/18
Answer: A.23/35
Explanation: Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that at least one green marble can be picked in the random draw of two marbles = Probability that one is green + Probability that both are green
= (⁶C₁ * ⁹C₁)/¹⁵C₂ + ⁶C₂/¹⁵C₂
= (6 * 9 * 2)/(15 * 14) + (6 * 5)/(15 * 14)
= 36/70 + 1/7
= 46/70 = 23/35
Answer:A
Question: If p and q are positive integers and q ≠ 1, then pq(q −2) is
A.pq^2
B.pq^2+2pq
C.pq^2-2pq
D.pq^2+pq
Answer: C.pq^2-2pq
Explanation: pq(q-2) = pq^2-2pq
Answer : C
Question: If a point (x,y) is randomly selected within the square shown in the figure above, (the vertices are on 1,1 1,-1, -1,1 -1,-1) what are
the W odds that x^2+y^2>1?
A.1. pi/4
B.2. W=pi/16
C.3. W=1-pi/4
D.4. W=1-pi/16
Answer: C.3. W=1-pi/4
Explanation: area of the square will be 4, and the area of circle with center (0,0) and radius 1 will be pi.
then succesful outcomes = 4 - pi (the area where x2+y2 will be greater than 1)
total outcomes = 4
therefore probability = (4-pi)/4
C
Question: It takes 60 identical printing machines 2 hours to print 200,000 cards. How long would it take 50 of these machines to print the same total?
A.2 hr 36 min
B.2 hr 40 min
C.2 hr 30 min
D.2 hr 25 min
Answer: A.2 hr 36 min
Explanation: 50% of 60 = 30% or 0.30
2 hr x 60 min = 120 min
120 min x 0.30 = 36 min
36 min + 2 hr = 2 hr and 36 min
Answer is A
Question: A, B and C invests Rs.2000, Rs.3000 and Rs.4000 in a business. After one year A removed his money; B and C continued the business for one more year. If the net profit after 2 years be Rs.3200, then A's share in the profit is?
A.288
B.266
C.299
D.400
Answer: D.400
Explanation: 2*12 : 3*12 : 4*24
1: 3: 4
1/8 * 3200 = 400
Answer:D
Question: A participated in cycling contest and he drove the lap at the rate of 6kmph, 12kmph, 18kmph, 24kmph..What is his average speed?
A.15kmph
B.18kmph
C.14.25 kmph
D.11.52 kmph
Answer: D.11.52 kmph
Explanation: =4x/(x/6 +x/12 + x/18 +x/24)
=11.52
ANSWER:D
Question: There are some sweets in party, 1/3rd eaten by kids . Then 2/5th of the remaining by men , then 2/3rd of the remaining by women . At last 2 were remaining . How many sweets were in total ?
A.45
B.27
C.15
D.26
Answer: C.15
Explanation: Sol:
If x sweets were there in total , then
x × (1 – 1/3)× (1 – 2/5) ×(1 – 2/3) = 2
x×2/3 × 3/5 × 1/3 = 2
x = 15
Answer:C
Question: 11 friends went to a hotel and decided to pay the bill amount equally. But 10 of them could pay Rs. 60 each as a result 11th has to pay Rs. 50 extra than his share. Find the amount paid by him.
A.110
B.111
C.115
D.123
Answer: C.115
Explanation: Explanation:
Average amount paid by 10 persons = Rs. 60
Increase in average due to Rs. 50 paid extra by the 11th men = Rs. 50/10= Rs. 5
Therefore, Average expenditure of 11 friends = Rs. 60 + Rs. 5 = Rs. 65
Therefore, Amount paid by the 11th men = Rs. 65 + Rs. 50 = Rs. 115
Correct Option: C
Question: The average of 1st 3 of 4 numbers is 6 and of the last 3 are 5. If the sum of the first and the last number is 17. What is the last numbers?
A.4
B.5
C.6
D.7
Answer: D.7
Explanation: A + B + C = 18
B + C + D = 15
A + D = 17
A – D = 3
A + D = 17
2D = 14
D = 7
ANSWER:D
Question: complete the numerical series with the correct number
6+3+2=181227
7+2+4=142840
8+6+3=482466
9+4+5=364567
6+5+5= ?
A.468645
B.303055
C.454585
D.454386
Answer: B.303055
Explanation: 6+5+5=303055
Answer : B
Question: a television have side 6*5.its screen size is 5*4.calculate the percentage of unscreened part t.v.
A.30
B.25
C.33.33
D.44.44
Answer: C.33.33
Explanation: unscreened part=30-20=10.
%unscreened part=10/30*100=33.33%
answer C
Question: 28 machines can do a work in 10days. How many machines are needed to complete the work in 40days?
A.10
B.6
C.4
D.7
Answer: D.7
Explanation: Required number of machines = 28*10/40 = 7
Answer is D