RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: A profit of Rs. 800 is divided between X and Y in the ratio of 1/2:1/3 . What is the difference between their profit shares ?
A.s. 260
B.s. 160
C.s. 360
D.s. 50
Answer: B.s. 160
Explanation: A profit of Rs. 800 is divided between X and Y in the ratio of 1/2 : 1/3 or 3:2.
so profits are 480 and 320.
difference in profit share = 480-320=160
ANSWER:B
Question: If each side of a square is increased by 25%, find the percentage change in its area.
A.46.25%
B.56.25%
C.56%
D.50.25%
Answer: B.56.25%
Explanation: Let each side of the square be a. Then, area = a2.
New side =(125a/100) =(5a/4). New area = (5a/4) 2=(25a2)/16.
Increase in area = ((25 a2)/16)-a2=(9a2)/16.
Increase% = [((9a2)/16)*(1/a2)*100] % = 56.25%.
ANSWER B 56.25%
Question: What day of the week was 7th January 1601
A.Sunday
B.Tuesday
C.Monday
D.Thursday
Answer: A.Sunday
Explanation: Explanation:
7th Jan 1601 = (1600 years + 7th Jan 1601)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)
7th Jan 1601 = 7 odd day
Total number of odd days = (0 + 7) = 7 = 0
0 odd days = Sunday
Hence 7th January 1601 is Sunday.
Answer: Option A
Question: In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State ?
A.2438
B.8000
C.2379
D.2988
Answer: B.8000
Explanation: Explanation:
Let the number of candidates appeared from each state be x.
In state A, 6% candidates got selected from the total appeared candidates
In state B, 7% candidates got selected from the total appeared candidates
But in State B, 80 more candidates got selected than State A
From these, it is clear that 1% of the total appeared candidates in State B = 80
=> total appeared candidates in State B = 80 x 100 = 8000
=> total appeared candidates in State A = total appeared candidates in State B = 8000
Answer: B) 8000
Question: A bag contains 20 red jellybeans and 20 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?
A.9/100
B.2/19
C.1/8
D.3/20
Answer: D.3/20
Explanation: Method-1
10 red jellybeans and 10 blue jellybeans
Total Outcomes = No. of ways to choose 3 Jelly bean at random out of a total 20 jellybeans = 20C3 = 1140
Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10C3 = 120
Probability = Favourable Outcomes / Total Outcomes = 10C3 / 20C3
Probability = 120 / 1140 = 2/19
Answer: option B
Method-2
Probability of First jelly bean to be Blue = 10/20 [Total 10 Blue out of total 20 jellybeans]
Probability of Second jelly bean to be Blue = 9/19 [Total 9 Blue remaining out of total 19 jellybeans remaining]
Probability of Third jelly bean to be Blue = 8/18 [Total 8 Blue remaining out of total 18 jellybeans remaining]
Required Probability = (10/20)*(9/19)*(8/18) = 3/20
Answer: option D
Question: If a, b, and c are distinct positive prime numbers, then (a^3)(b)(c) has how many different positive factors greater than 1?
A.a. 3
B.b. 9
C.c. 12
D.d. 15
Answer: A.a. 3
Explanation: Let Number is (a^3)(b)(c) = (1^3)(2)(3) = 6
We can write 6 as product of two numbers in following ways
1*6
2*3
2 cases = 2*2 i.e. 4 factors (including 1)
Factors greater than 1 = 3
Answer: Option A
Question: A trader sold an item at a loss of 18%. if he had increased the price by $25 he would have made a gain of 5%. what is the cost price of the item?
A.$414.78
B.$424.78
C.$534.78
D.$434.78
Answer: D.$434.78
Explanation: Let C.P. be $x
Then 105% of x - 82% of x = 100
23% of x = 100
23/100 *X = 100
23x = 10000
x = 434.78
Answer is D
Question: There are 12 pieces of radioactive metal W that look identical. 11 of the pieces give the same radiation count when measured, the 12th piece is a counterfeit and gives a different radiation level, which may be more or less than the other 11. We are given a radiation scale, which can take 2 sets of samples and compare their added up radiation levels to tell us if the sums are the same or if different, which set has the higher level of radiation. What is the minimum number of comparisons we need on this scale to identify the counterfeit sample and to also determine whether it has more or less radiation than the other samples ?
A.2
B.3
C.4
D.5
Answer: B.3
Explanation: First of all if you are down to just 3 pieces and you know that if the offending piece is less or more active, then it takes exactly 1 measurement to find out the offending piece. So you know you have to reduce the problem to three.
Now when you are down to either A or B after measurement 1, you need the next measurement to (a) reduce the problem set to 3 and (b) to know whether anser is more or less. Now you cannot compare a group of 4 to 4, as in the best case it will only reduce the problem to 4 elements which is not good enough.
If you have to choose a set of 3 to compare, you cannot pick any 3 on the same side from the same set (A or B) because if you do this, a quick check will show you that in every choice there is a case where you can only get down to 4 elements. Eg. If you weighed {1,2,3} v/s {5,9,10} and they were equal you're problem would only reduce to {4,6,7,8}
The easiest way to solve this then is to compare 3 to 3, and make sure each side has elements from both AB such that whatever the measurement outcome in the worst case the problem reduces to 3 elements only. Which is why the sets {1,5,9} and {2,6,7} OR {A,B,C}{A,B,B}. The extra element from C is just taken to make the problem symmetric so to say, we have 8 elements and we make it 9, to compose 3 sets of 3 each.=B
Question: Of 70 players on a football team, 28 are throwers. The rest of the team is divided so one third are left- handed and the rest are right handed. Assuming that all throwers are right handed, how many right- handed players are there total?
A.56
B.59
C.63
D.71
Answer: A.56
Explanation: Total = 70
Thrower = 28
Rest = 70 - 28 = 42
Left Handed = 42/3 = 14
Right handed = 28
If all Thrower are right handed then total right handed is 28 + 28 = 56
so A. 56 is the right answer
Question: Company X sells a selection of products at various price points. Listed below are unit sales made for one particular day. How many unit sales on that day were greater than the mean sale price but less than the median sale price?
$50, $50, $97, $97, $97, $120, $125, $155, $155, $199, $199
A.0
B.1
C.2
D.3
Answer: B.1
Explanation: Taking the prices of products in ascending order(already arranged)
$50, $50, $97, $97, $97, $120, $125, $155, $199, $199, $199
we see that median value is the 6th value as there in total 11 values given
Arithmetic mean = total/number of entries = 1388/11 =126.1818
We are asked to find how many unit sales on that day were greater than the mean sale price but less than the median sale price
as we can clearly see that there is one value between $120 and $126.1818, answer is 1 unit
Correct Answer - B
Question: A car takes 6 hours to cover a distance of 540 Km. how much should the speed in Kmph be maintained to cover the same direction in 3/2th of the previous time?
A.50 Kmph
B.60 Kmph
C.70 Kmph
D.80 Kmph
Answer: B.60 Kmph
Explanation: Time = 6
Distance = 540
3/2 of 6 hours = 6 * 3/2 = 9 Hours
Required speed = 540/9 = 60 Kmph
B
Question: In an examination, 30% of total students failed in Hindi, 35% failed in English and 35% in both. The percentage of these who passed in both the subjects is :
A.10%
B.20%
C.30%
D.70%
Answer: D.70%
Explanation: pass percentage = 100 - (30+35-35) = 100-30 = 70
ANSWER:D
Question: Find the invalid no.from the following series 15 46 63 71 75 77 78
A.15
B.46
C.63
D.71
Answer: B.46
Explanation: In this question, if you start from the beginning, firstly the difference is 31, then the difference is 17, then the difference is 8. So there is no logic. If you start backwards,you can see that firstly, 1 is added, then 2 is added, then 4 is added, then 8,16 and 32 should have been added. So you have got the right clue. So 15 + 32 the first number should be 47 and afterwards, the series becomes + 16, + 8, + 4and so on.
46
B
Question: At the end of a business conference the ten people present all shake hands with each other once. How many handshakes will there be altogether ?
A.20
B.45
C.55
D.90
Answer: B.45
Explanation: Explanation:
Clearly, total number of handshakes = (9+ 8 + 7 + 6 + 5 + 4 + 3 + 2+1) = 45.
Answer: B
Question: The probability of having a girl is identical to the probability of having a boy. In a family with five children, what is the probability that all the children are of the same gender?
A.1/16
B.1/4
C.1/8
D.1/12
Answer: A.1/16
Explanation: There are two cases. All boys and all girls
case 1. all boys.
1/2*1/2*1/2*1/2*1/2= 1/32
case 2 .all girls.
1/2*1/2*1/2*1/2*1/2=1/32
We will add the cases to get 1/16
Question: In a school 50% of the students are younger than 10, 1/20 are 10 years old and 1/10 are older than 10 but younger than 12, the remaining 70 students are 12 years or older. How many students are 10 years old?
A.5
B.10
C.15
D.20
Answer: B.10
Explanation: Solution
Let us write the fraction for each group of students
Group A: younger than 10: 50% = 50/100 = 1/2
Group B: 10 years old: 1/20
Group C: older that 10 but younger than 12: 1/10
Group D: 12 years or older: 70 students
The fraction for group A, B and C together is given by
1 / 2 + 1 / 20 + 1 / 10 = 10 / 20 + 1 / 20 + 2 / 20 , common denominator
= 13 / 20 , add numerators
The fraction for group D is given by
20 / 20 - 13 / 20 = 7 / 20 and corresponds to 70 students
If X is the total number of students, then
7 / 20 of X = 70
or (7 / 20) × X = 70
Solve for X
X = 70 × (20 / 7) = 200
Students who are 10 years old have a fraction of 1 / 20 of the total X and their number is equal to
(1 / 20) of 200 = 10 students
ANSWER IS B
Question: How many digits are in (8×10^10)(10×10^5)?
A.17
B.25
C.26
D.27
Answer: A.17
Explanation: The question simplfies to (8×10^10)(10^6)
=> 8*10^16
=> Will contain 16 zeros +1 digit 8
=>17
Ans A
Question: Set A is {1, 3, 5, 7, 9}
Set B is {16, 18, 20, 22, 24}
For the sets of numbers above, which of the following is true?
I. The mean of Set B is greater than the mean of Set A.
II. The median of Set B is greater than the median of Set A.
III. The standard deviation of Set B is greater than the standard deviation of Set A.
A.I only
B.I and II only
C.I and III only
D.II and III only
Answer: B.I and II only
Explanation: The mean and median of Set A is 5.
The mean and median of Set B is 20.
Thus statements I and II are both true.
The standard deviation is related to the distance from the mean.
Since the elements of both sets are spaced identically around the mean (+/- 2, +/- 4), the standard deviations of both sets are equal.
The answer is B.
Question: In the first half of the 20th century, the population of a particular country increased by 200 percent. In the second half of the century, the population increased by 300 percent. What was the percent increase for the 20th century as a whole?
A.500%
B.600%
C.800%
D.1100%
Answer: D.1100%
Explanation: Say initially population was 100.
What is 200% of 100? It is 200/100 * 100 = 200.
An increase of 200% means the new population became 100 + 200 = 300
What is 300% of 300? It is 300/100 * 300 = 900
An increase of 300% means the new population now is 300 + 900 = 1200
So from 100, the population increased to 1200 i.e. an increase of 1100.
1100 is what percent of 100?
1100 = x/100 * 100
i.e. it is 1100%
ANSWER:D
Question: A vendor bought Headphones at 5 for Rs 100.How many for Rs 100 must he sell to gain 50%?
A.8
B.3
C.5
D.6
Answer: D.6
Explanation: C.P. of 9 Headphones= Rs100
S.P. of 9 Headphones = 150% of Rs 100 = Rs150
For Rs.150 , Headphones sold = 9.
For Rs. 100,Pens sold=(9/150)*100=6
Answer: Option D
Question: (500 + 200+100) Ã— 4 Ã— (3 + 4) = ?
A.52000
B.32400
C.22400
D.22800
Answer: C.22400
Explanation: (500 + 200+100) Ã— 4 Ã— (3 + 4) = ?
or, ? = 800 Ã— 4 Ã— 7 = 22400
Answer C
Question: P is four times as old as Q, and Q is six years older than R. If R is z years old, what is P’s age in terms of z?
A.4z + 6
B.4z − 24
C.4z + 24
D.(z + 6)/4
Answer: C.4z + 24
Explanation: P = 4Q
Q = R+6 = z+6
P = 4(z+6) = 4z + 24
The answer is C.
Question: By the how much is two-fifth of 400 greater than three -fifths of 150?
A.15
B.70
C.5
D.30
Answer: B.70
Explanation: Reqd no. = 2â„5 Ã— 400 - 3â„5 Ã— 150
= 160 - 90 = 70
Answer B
Question: Two cars start from the opposite places of a main road, 150 km apart. First car runs for 30 km and takes a right turn and then runs 15 km. It then turns left and then runs for another 25 km and then takes the direction back to reach the main road. In the mean time, due to minor break down the other car has run only 35 km along the main road. What would be the distance between two cars at this point?
A.65
B.38
C.60
D.28
Answer: C.60
Explanation: Answer: C) 60 km
Question: If the sales tax reduced from 3 1/2 % to 3 1/3%, then what difference does it make to a person who purchases an article with market price of Rs. 8400 ?
A.11
B.13
C.14
D.18
Answer: C.14
Explanation: Required difference = [3 ½ % of Rs.8400] – [3 1/3 % of Rs.8400]
= [(7/20-(10/3)]% of Rs.8400 =1/6 % of Rs.8400
= Rs. [(1/6)8(1/100)*8400] = Rs. 14.
Answer is C.
Question: P is four times as fast as Q and working together, they can complete a work in 12 days. In how many days can Q alone complete the work?
A.35 days
B.15 days
C.16 days
D.25 days
Answer: B.15 days
Explanation: P = 4Q
P + Q = 4Q + Q = 5Q
These 5Q people can do the work in 12 days, which means Q can do the work in 60 days.
Hence, P can do the work in 15 days.
ANSWER:B
Question: Find the average of the series : 150 , 160, 328, 147 and 120?
A.174
B.178
C.181
D.180
Answer: C.181
Explanation: Average = (150 + 160 + 328 + 147 + 120)/5
= 905/5
= 181
Answer: C
Question: Of the three numbers, second is twice the first and is also thrice the third.If the average of the three numbers is 44.Find the largest number.
A.42
B.48
C.72
D.67
Answer: C.72
Explanation: Sol: Let the third number be x.
Then second number = 3x.
First number=3x/2.
Therefore x+3x+(3x/2)=(44*3) or x=24
So largest number= 2nd number=3x=72
ANSWER :C
Question: The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team ?
A.11 year
B.12 year
C.23 year
D.25 year
Answer: C.23 year
Explanation: Let the average age of the whole team be x years.
11x - (26 + 29) = 9 (x - 1)
=> 11x - 9x = 46
=> 2x = 46
=> x = 23.
So, average age of the team is 23 years.
Answer: C
Question: The average of 9 observations was 7, that of the 1st of 5 being 10 and that of the last 5 being 8. What was the 5th observation?
A.27
B.12
C.15
D.17
Answer: A.27
Explanation: Explanation:
1 to 9 = 9 * 7 = 63
1 to 5 = 5 * 10 = 50
5 to 9 = 5 * 8 = 40
5th = 50 + 40 = 90 – 63 = 27
Option A
Question: In a hotel, 60% had vegetarian lunch while 30% had non-vegetarian lunch and 15% had both type of lunch. If 96 people were present, how many did not eat either type of lunch ?
A.21
B.22
C.23
D.24
Answer: D.24
Explanation: n(A)=(60/100 ∗96)=288/5
n(B)=(30/100 ∗96)=144/
n(A∩B)=(15/100 ∗96)=72/5
People who have either or both lunchn(A∪B)=288/5 +144/5 −72/5 =360/5 =72
So People who do no have either lunch were = 96 -72 = 24 ANSWER : D
Question: Which one of the following numbers is divisible by 99?
A.3572404
B.135792
C.913464
D.114345
Answer: D.114345
Explanation: Explanation:
The sum of all the digits of 114345 is 18.
The sum of all digits at even places and odd places are 9 and 9 respectively.
The difference between the sums is 0.
So, the number is divisible by both 9 and 11 and hence by 99.
Answer:D
Question: In a circular track of length 100 m, three persons A, B and C start together. A and B start in the same direction at speeds of 10 m/s and 8 m/s respectively. While C runs in the opposite at 15 m/s. when will all the three meet for the first time after the start?
A.90
B.100
C.110
D.120
Answer: B.100
Explanation: A takes (100m) / (10m/s) = 10s to finish one complte round.
B takes (100m) / (8m/s) = 12.5s to finish one complte round.
C takes (100m) / (15m/s) = (20/3)s to finish one complte round.
For them to meet, the time will be LCM of above three times ie LCM(10,12.5,20/3)
= LCM(10/1,25/2,20/3)
= LCM(10,25,20)/GCD(1,2,3)
=100/1
=100
ANSWER:B
Question: Johnson has a corporate proposal. The probability that vice-president Adams will approve the proposal is 0.3. If VP Adams approves the proposal, then the probability that vice-president Baker will approve the proposal is 0.4. If VP Adams doesn’t approve the proposal, then the probability that vice-president Baker will approve the proposal is 0.3. What is the probability that at least one of the two VPs, approves Johnson’s proposal?
A.0.12
B.0.42
C.0.28
D.0.48
Answer: B.0.42
Explanation: (1at approves * second rejects ) + (1st rejects * 2nd approves ) + (both approves )
=0.3*0.2+0.4*0.3+0.6*0.4
= 0.06+0.12+0.24
=0.42
B
Question: A can do a piece of work in 16 days and B can do it in 15 days and C can do it 20 days. They started the work together and A leaves after 2 days and B leaves after 4 days from the beginning. How long will work lost?
A.10 2/5
B.10 2/0
C.12 1/6
D.10 2/1
Answer: C.12 1/6
Explanation: 2/16 + 4/15 + x/20 = 1
x = 73/6 = 12 1/6 Answer: C
Question: If the price of a computer were reduced by 12 percent, which of the following could not be the final price of the computer? (Assume initial price to be integer in cents)
A.$844.10
B.$895.40
C.$1,053.80
D.$1,135.20
Answer: A.$844.10
Explanation: Let X be the initial price of computer without discount
then price (final) after discount should be=X(1-12/100)------->X(22/25)=A(say)
means X=A*(25/22).....
so initial price to be integer(As per stem) final price must be multiple of 22
if we check options all were divisible by 22 except option (A)..which is Ans.
Ans A
Question: 15 buses are running between two places P and Q. In how many ways can a person go from P to Q and return by a different bus?
A.210
B.320
C.150
D.220
Answer: A.210
Explanation: we can go in any bus out of the 15 buses.
Hence He can go in 15 ways.
Since we can not come back in the same bus that he used for travelling,
we can return in 14 ways.
Total number of ways = 15 x 14 = 210
ANSWER:A
Question: The total number of plums that grow during each year on a certain plum tree is equal to the number of plums that grew during the previous year, less the age of the tree in years (rounded down to the nearest integer). During its 5th year, the plum tree grew 50 plums. If this trend continues, how many plums will it grow during its 6th year?
A.40
B.42
C.45
D.48
Answer: C.45
Explanation: 1st year : 0-1 (age), we take age = 0 (as the question says that we have to(rounded down to the nearest integer))
2ndyear : 1-2 (age), we take age = 1
3rd year : 2-3 (age), we take age = 2
4th year : 3-4 (age), we take age = 3
5th year : 4-5 (age), we take age = 4
6th year : 5-6 (age), we take age = 5
Thus for the 5th year = 50,
6th year = 50-5 = 45.
The correct answer is C.
Question: On a certain test, Bill scored 20 more points than John, but half as many points as Sue. If the sum of the scores of the three students was 200 points, how many points did Bill receive?
A.30
B.40
C.50
D.55
Answer: D.55
Explanation: consider john's points as x
so bill's points are x +20
So sue's points are 2x +40
total of all three students is 4x +60, but this total is given as 200
x = 35
so bill's points = 35 +20 = 55
correct answer option D
Question: What is the total number of integers between 100 and 250 that are divisible by 5?
A.27
B.28
C.29
D.30
Answer: C.29
Explanation: 105, 110, 115, ..., 240,245
This is an equally spaced list; you can use the formula:
n = (largest - smallest) / ('space') + 1 = (245 - 105) / (5) + 1 = 140/5 + 1 = 28 + 1 = 29
Answer is C
Question: Evaluate permutation
4P4
A.22
B.23
C.24
D.25
Answer: C.24
Explanation: Explanation:
nPn=n!
4P4=4∗3∗2∗1=24
Option C
Question: The ratio between the present ages of A and B is 5:3 respectively. The ratio between A's age 4 years ago and B's age 4 years hence is 1:1. What is the ratio between A's age 4 years hence and B's age 4 years ago?
A.3:9
B.3:6
C.3:1
D.3:4
Answer: C.3:1
Explanation: Let the present ages of A and B be 5x and 3x years respectively.
Then, (5x - 4)/(3x + 4) = 1/1
2x = 8 => x = 4
Required ratio = (5x + 4):(3x - 4) = 24:8 = 3:1
Answer: C
Question: The ratio between the sale price and the cost price of an article is 10:7. What is the ratio between the profit and the cost price of that article?
A.2:9
B.2:5
C.3:7
D.2:0
Answer: C.3:7
Explanation: Let C.P. = Rs. 7x and S.P. = Rs. 10x.
Then, Gain = Rs. 3x
Required ratio =3x : 7x = 3:7
Answer:C
Question: A man purchases some eggs for Rs.500. 15% of the eggs break during transportation. What profit percentage should he take on the remaining to get an overall profit of 20%?
A.37.5
B.41.17
C.21.17
D.45.17
Answer: B.41.17
Explanation: Explanation:
The eggs remaining = 85%
CP of 85% eggs = 85x500/100 = Rs.425
Overall profit = 500 + 20% of 500 -CP of 85% eggs.
Rs.600 – Rs.425 = Rs.175.
:. Overall profit percent = 175x100/425 = 41.17%.
Answer: Option B
Question: Two employees X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?
A.Rs. 150
B.Rs. 300
C.Rs. 250
D.Rs. 200
Answer: C.Rs. 250
Explanation: Explanation:
Let the amount paid to X per week =x
and the amount paid to Y per week =y
12y/10+y=550
22y/10=550
y=5500/22=500/2=250
ANSWER IS C
Question: How many three digit numbers X greater than 330 can be formed from the digits 1, 2, 3, 4, 5 and 6, if each digit can only be used once?
A.72
B.48
C.120
D.96
Answer: A.72
Explanation: Find the number of 3 digit numbers greater than 300, and then subtract from them the number of three digit numbers between 300 and 330 formed with the given 6 digits.
ABC three digit numbers greater than 300 - total of 4*5*4=80, because a must be greater than 2, so 4 possibilities left; B 5 remaining, and C 4 remaining possibilities to choose from.
Between 300 and 330 there are 1*2*4=8 numbers with the given property: A = 3, 1 possibility; B can be only 1 or 2 (ABC < 330), so two possibilities; C the remaining 4 possibilities after choosing A and B.
Total possible choice 80 - 8 =72=X
Answer: A.
Question: A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 30 seconds. What is the length of the goods train?
A.230 m
B.350 m
C.260 m
D.270 m
Answer: B.350 m
Explanation: Explanation:
Speed =[ 72 x (5/18) ]m/sec= 20 m/sec.
Time = 30 sec.
Let the length of the train be x metres.
Then,[ (x+250)/30 ]= 20
=> x + 250 = 600
=> x = 350. Answer: B
Question: A 180 meter long train crosses a man standing on the platform in 6 sec. What is the speed of the train?
A.107 kmph
B.138 kmph
C.121 kmph
D.108 kmph
Answer: D.108 kmph
Explanation: D
108 kmph
S = 180/6 * 18/5 = 108 kmph
Question: An amount at compound interest sums to Rs.17640/- in 2 years and to Rs.19404/- in 3 years at the same rate of interest. Find the rate percentage?
A.5%
B.7%
C.9%
D.10%
Answer: D.10%
Explanation: Explanation:
The difference of two successive amounts must be the simple interest in 1 year on the lower amount of money.
S.I = 19404/- - 17640/- = Rs. 1764/-
Rate of interest = (1764/17640) × (100/1)
=>10%
Answer: Option D
Question: Two trains are running in opposite directions with the same speed. If the length of each train is 120 metres and they cross each other in 12 seconds, then the speed of each train (in km/hr) is:
A.10
B.18
C.36
D.72
Answer: C.36
Explanation: Explanation:
Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec.
So, 2x =	(120 + 120) /12
2x = 20
x = 10.
Speed of each train = 10 m/sec = 10 x	18/5 km/hr = 36 km/hr.
ANSWER IS C