RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: In the youth summer village there are 125 people, 50 of them are not working, 25 of them have families and 75 of them like to sing in the shower. What is the largest possible number of people in the village, which are working, that don't have families and that are singing in the shower?
A.75
B.80
C.85
D.100
Answer: A.75
Explanation: total =125
not working =50
having family=25
like to sing in shower = 75
working=125-50=75
not having family = 125-25=100
like to sing in shower = 100
largest possible number is the lowest possible among the above
Thus 75
A
Question: Average of five consecutive even numbers is 35. Find the greatest number in these five numbers?
A.37
B.38
C.39
D.41
Answer: C.39
Explanation: 39
Option 'C'
Question: The compound interest accrued on an amount of Rs.44000 at the end of two years is Rs.1193.60. What would be the simple interest accrued on the same amount at the same rate in the same period?
A.Rs.10,840
B.Rs.10,720
C.Rs.10,560
D.Rs.10,280
Answer: C.Rs.10,560
Explanation: Let the rate of interest be R% p.a.
4400{[1 + R/100]2 - 1} = 11193.60
[1 + R/100]2 = (44000 + 11193.60)/44000
[1 + R/100]2 = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]2 = 784/625 = (28/25)2
1 + R/100 = 28/25
R/100 = 3/25
Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100
=Rs.10560
ANSWER:C
Question: How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)(5!)(6!)(7!)(8!)(9!)?
A.504
B.672
C.864
D.936
Answer: B.672
Explanation: N=(1!)(2!)(3!)(4!)(5!)(6!)(7!)(8!)(9!)
1!=1
2!=1*2
3!=1*2*3
4!=1*2*3*4
5!=1*2*3*4*5
6!=1*2*3*4*5*6
7!=1*2*3*4*5*6*7
8!=1*2*3*4*5*6*7*8
9!=1*2*3*4*5*6*7*8*9
N=(2^8)(3^7)(4^6)(5^5)(6^4)(7^3)(8^2)(9^1)
=(2^30)(3^13)(5^5)(7^3)
To find the number of perfect squares each exponent in the prime factorization must be even .
Comparing to the form (2^p)(3^q)(5^r)(7^s).
In this case,
p can take values from 0 to 30
q can take values from 0 to 13
r can take values from 0 to 5
s can take values from 0 to 3
Number of even exponents
p=16
q=7
r=3
s=2
Number of perfect squares= 16*7*3*2 = 672
Answer B
Question: A rectangular tiled patio is composed of 48 square tiles. The rectangular patio will be rearranged so that there will be 2 fewer columns of tiles and 4 more rows of tiles. After the change in layout, the patio will still have 48 tiles, and it will still be rectangular. How many rows are in the tile patio before the change in layout?
A.5
B.6
C.10
D.13
Answer: B.6
Explanation: Suppose there are c columns and there are r rows
Original Situation
So, Number of tiles = c*r = 48
Also. Reach column has r tiles and each row has c tiles
New Situation
Number of tiles in each column is r-2 and number of tiles in each row is c+4
So, number of rows = r-2 and number of columns is c+4
So, Number of tiles = (r-2)*(c+4) = 48
Comparing both of them we get
c*r = (r-2)*(c+4)
=> 4r -2c = 8
c = 2r - 4
Putting it in c*r=48
(2r-4)*r = 48
2r^2 - 4r - 48=0
r cannot be negative so r = 6
and c = 8
So, Answer will be B
Question: 7 people, A, B, C, D, E, F and H, go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. In how many different arrangements will there be at least one person between A and F?
A.8! - 7!2
B.5! - 7!2
C.6! - 7!2
D.8! - 6!2
Answer: A.8! - 7!2
Explanation: We need to find at least one person between A and F which is equal to Total ways - No one between A and F
Eight people can be arranged in 8! ways.
No one between A and F is equal to 7!*2! (consider A and F as one. 7 people can be arranged in 7! and A and F in 2!)
So, total ways = 8! - 7!2
ANS:A
Question: A man and a boy complete a work together in 24 days. If for the last 6 days man alone does the work then it is completed in 26 days. How long the boy will take to complete the work alone?
A.72 days
B.20 days
C.24 days
D.36 days
Answer: A.72 days
Explanation: Explanation:
(man + boy)’s 1 day’s work = 1/24
Their 20 day’s work = 1/24 × 20 = 5/6
The remaining 1/6 work is done by the man in 6days
Therefore, the man alone will finish the work in 6 × 6 days = 36 days
Man’s 1 day’s work = 1/36
Therefore, boy’s 1 day’s work = 1/24 – 1/36 = 3 – 2 /72 = 1/72
Therefore, the Boy alone will finish the work in 72 days.
Answer: Option A
Question: Albert is 2 times Mary’s age and 4 times as old as Betty. Mary is 22 years younger than Albert. How old is Betty?
A.8
B.9
C.11
D.15
Answer: C.11
Explanation: A = 2M = M+22
M = 22
A = 44
A = 4B, and so B = 11
The answer is C.
Question: The average (arithmetic mean) of four numbers is 4x + 1. If one of the numbers is x, what is the average of the other three numbers?
A.x + 1
B.3x + 3
C.5x + 4/3
D.5x + 4
Answer: C.5x + 4/3
Explanation: If the average of four numbers is 4x + 1, then the sum of all four numbers if 4 * (4x + 1) = 16x + 12. If one of the numbers is x, then the average of the remaining three numbers would be: (16x + 4 - x)/3 = (15x + 4)/3 = 5x + 4/3.
Answer is C.
Question: At a certain amateur bands rock show the ratio of freaks to nerds (N) to geeks is 1:2:3. If these are the only types of people in the crowd, and each person in the crowd has the same chance of yelling rock on!, what are the chances that the next person who yells rock on!will be a geek?
A.1
B.1/2
C.1/3
D.1/4
Answer: B.1/2
Explanation: Another approach is toassign some nice valuesfor the number of freaks, nerds and geeks.
Given: the ratio of freaks to nerds to geeks is 1:2:3
So, let's say there is: 1 freak, 2 nerds and3geeks for a TOTAL of6people.
P(next person who yellsrock on!is a geek) = (# of geeks in the crowd)/(TOTAL # of people in the crowd)
=3/6
= 1/2
Answer:
B
Question: What is the product of all the possible values of x if x^2 - 9x + 20?
A.–29
B.–12
C.12
D.20
Answer: D.20
Explanation: Explanation:
=> y= x^2 - 9x + 20
=> y= (x-4) (x-5)
=> x= 4, y= 5
Product y = (4)(5) = 20
answer option 20
ANSWER:D
Question: Which of the following number is divisible by 24 ?
A.18
B.19
C.20
D.24
Answer: D.24
Explanation: 24 = 3 x8, where 3 and 8 co-prime.
Clearly, 35718 is not divisible by 8, as 718 is not divisible by 8.
Similarly, 63810 is not divisible by 8 and 537804 is not divisible by 8.
Consider option (D),
Sum of digits = (3 + 1 + 2 + 5 + 7 + 3 + 6) = 27, which is divisible by 3.
Also, 736 is divisible by 8.
3125736 is divisible by (3 x 8), i.e., 24.
D)
Question: An arithemetic sequence of numbers begins p, 2p + 2, 4p + 3... What is the 10th member of this sequence?
A.A)37
B.B)28
C.+ 13
D.D)53
Answer: B.B)28
Explanation: Common difference of an arithmetic series is constant.
2p+2-p=4p+3-2p-2
p=1
1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37
Ans:B
Question: A one-foot stick is marked in 1/4 and 1/6 portion. How many total markings will there be, including the end points?
A.8
B.10
C.12
D.14
Answer: A.8
Explanation: LCM of 24 = 12
1/4 marking are (Table of 3) 0...... 3 ........... 6.......... 9 .......... 12
(Total = 5)
1/6 marking are (Table of 2) 0 ....... 2 ...... 4...... 6........ 8......... 12 ((Total = 6)
Overlapping markings are 0 ........ 6 ......... 12 (Total = 3)
Total markings = 5+6-3 = 8
Answer = A
Question: Find the odd one out:
125, 106, 87, 76, 65, 58, 53
A.33
B.87
C.26
D.18
Answer: B.87
Explanation: Explanation:
This sequence represents a series in which from the reverse order a prime number is added:
53+5=58
58+7=65
65+11=76
76+13=89
89+17=106
106+19=125
Hence 87 is the answer.
ANSWER: B
Question: When running a mile during a recent track meet, Nuria was initially credited with a final time of 5 minutes, 30 seconds. Shortly after her run, officials realized that the timing mechanism malfunctioned. The stopwatch did not begin timing her until 2/7 of a minute after she began to run. If the time was otherwise correct, how long did it actually take Nuria to run the mile?
A.5 minutes, 17.6 seconds
B.5 minutes, 21.8 seconds
C.5 minutes, 23.6 seconds
D.5 minutes, 44.44 seconds
Answer: D.5 minutes, 44.44 seconds
Explanation: One approach:
The watch starts to work after Nuria began his running. It means the time should be greater than credited 5 minutes, 30 seconds. The only number is 5 minutes, 44.44 seconds.
Another approach:
2/7 close to 20 second when added to the 5 minutes, 30 seconds, it means it passes 5 minute 44.44 seconds
Answer: D
Question: In a group of 400 readers who read science fiction or literacy works or both, 250 read science fiction and 230 read literacy works. How many read both science fiction and literacy works?
A.70
B.80
C.90
D.100
Answer: B.80
Explanation: Consider total number of reader n(S U L) = 400
People who read science fiction n(S) = 250
People who read literacy works n(L) = 230
both science fiction and literacy n(S ∩ L) = ?
n(S U L) = n(S)+ n(L) - n(S ∩ L)
400 = 250 + 230 - n(S ∩ L)
n(S ∩ L) = 480 - 400
n(S ∩ L) = 80
SO People read both science fiction and literacy works are 80
ANSWER:B
Question: Lindsay can paint 1/x of a certain room in 25 minutes. What fraction of the same room can Joseph paint in 25 minutes if the two of them can paint the room in an hour, working together at their respective rates?
A.1/7x
B.3x/(x – 3)
C.(3x – 7) / 7x
D.1/ (x – 3)
Answer: C.(3x – 7) / 7x
Explanation: Lindsey and Joseph together can paint the room in 1 hour
in 25 mins they can paint 3/7 part of the room.
Lindsay alone paints 1/x part of the room in 25 mins
Take Joseph paints 1/y parts in 25 mins
1/x + 1/y = 3/7
1/y = 3/7 - 1/x = (3x-7)/7x ANs C
Question: 5 girls and 3 boys are arranged randomly in a row. Find the probability that:
1) there is one boy on each end.
2) There is one girl on each end.
A.3/28 & 5/14
B.5/28 & 3/14
C.5/14 & 3/28
D.3/28 & 1/14
Answer: A.3/28 & 5/14
Explanation: Total # of arrangements is 8!;
1. The # of arrangements where there is one boy on each end is C^2/3∗2∗6! where C^2/3C32 is # of ways to select which 2 boys will be on that positions, *2 # of way to arrange them on that position (A------B and B------A) and 6! is # of arrangements of 6 people left;
P=C^2/3∗2∗6!/8!=3/28
2. The # of arrangements where there is one girl on each end is C^2/5∗2∗7!, where C^2/5 is # of ways to select which 2 girls will be on that positions, *2 # of way to arrange them on that position (A------B and B------A) and 6! is # of arrangements of 6 people left;
P=C^2/5∗2∗6!/8!=5/14
Answer: A.
Question: A Paul's company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 percent from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996?
A. 1:2
B. 4:5
C. 1:1
D. 3:2
Answer: A. 1:2
Explanation: A Paul's company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996.1996- x----- y------- x+y
1997- x1---- y1-- x1+y1
x1= a*x .....(P)
y1= 1.07*y...(Q)
x1+y1/x+y = 1.01
solve using P and Q and find x/y
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.2787
B.288
C.277
D.400
Answer: D.400
Explanation: Explanation:
2*12 : 3*12 : 4*24
1: 3: 4
1/8 * 3200 = 400
Answer: D
Question: On dividing 73 by a number, the quotient is 9 and the remainder is 1. Find the divisor?
A.A)8
B.B)15
C.C)16
D.D)17
Answer: A.A)8
Explanation: d = (D-R)/Q
= (73 - 1)/9
= 72/9 = 8
A)
Question: In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?
A.30%
B.70%
C.100%
D.120%
Answer: B.70%
Explanation: Solution
Let C.P. = Rs. 100.	Then, Profit = Rs.320,
S.P. = 420.
Let C.P. = Rs. 125% of Rs. 100	= Rs.125.
New S.P. = Rs.420.
Profit = Rs. (420 - 125)	= Rs.295.
∴ Required percentage	= (295 / 420 x 100)%
= 1475 / 21%
= 70%.
Answer B
Question: A batsman makes a score of 92 runs in the 17th inning and thus increases his average by 3. Find his average after 17th inning.
A.36
B.39
C.42
D.44
Answer: D.44
Explanation: Let the average after 17th inning = x.
Then, average after 16th inning = (x – 3).
∴ 16(x – 3) + 92 = 17x or x = (92 – 48) = 44.
Answer D
Question: Using all the letters of the word "PROBLEMS", how many words can be formed, which begin with P and end with S?
A.680
B.560
C.720
D.120
Answer: C.720
Explanation: There are eight letters in the given word.
Consider 8 blanks ....
The first blank and last blank must be filled with P and S all the remaining six blanks can be filled with the remaining 6 letters in 6! ways.
The number of words = 6! = 720.
ANSWER:C
Question: A bag contains 2 yellow, 5 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
A.7/12
B.5/12
C.3/7
D.2/7
Answer: A.7/12
Explanation: Total number of balls = 2 + 5 + 2 = 9
Let S be the sample space.
n(S) = Total number of ways of drawing 2 balls out of 9 = 9C2
Let E = Event of drawing 2 balls , none of them is blue.
n(E) = Number of ways of drawing 2 balls , none of them is blue
= Number of ways of drawing 2 balls from the total 7 (=9-2) balls = 7C2
(∵ There are two blue balls in the total 9 balls. Total number of non-blue balls = 9 - 2 = 7)
P(E) = n(E)/n(S)
=7C2 / 9C2
=7/12
Ans- A
Question: 896.7 – 573.07 – 95.007 = ?
A.228.623
B.224.777
C.233.523
D.414.637
Answer: A.228.623
Explanation: Solution
Given expression = 896.7 - (573.07 + 95.007) = 896.7 - 668.077 = 228.623. Answer A
Question: Two cars are traveling in the same direction along the same route. The red car travels at a constant speed of 10 miles per hour, and the black car is traveling at a constant speed of 50 miles per hour. If the red car is 20 miles ahead of the black car, how many hours will it take the black car to overtake the red car?
A.0.1
B.0.5
C.1
D.1.2
Answer: B.0.5
Explanation: Option B
20+10t=50t
t=0.5
Question: 20 workers can finish a work in 30 days. After how many days should 5 workers leave the job so the work is completed in 35 days?
A.7
B.6
C.15
D.4
Answer: C.15
Explanation: Work done by 15 workers in 35 days = 15*35=525 men days.
Now, work remaining= 600–525=75 men days and we have 5 workers .
So 5 workers will work for 75/5=15 days.
ANSWER:C
Question: Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand's present age in years?
A.24
B.89
C.67
D.56
Answer: A.24
Explanation: Let the present ages of Sameer and Anand be 5x years and 4x years respectively.
Then,	(5x + 3/4x + 3)=11/9
9(5x + 3) = 11(4x + 3)
45x + 27 = 44x + 33
45x - 44x = 33 - 27
x = 6.
Anand's present age = 4x = 24 years.
Answer:A
Question: A man has Rs. 80 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?
A.7
B.2
C.13
D.15
Answer: D.15
Explanation: Let number of notes of each denomination be x.
Then, x + 5x + 10x = 80 16x = 80 x = 5.
Hence, total number of notes = 3x = 15.
Answer:D
Question: Solve for the given equation 121 x 54 =?
A.68225
B.75625
C.76569
D.45854
Answer: B.75625
Explanation: Explanation:
This question can be solved by 2 approaches:
=> Direct Multiplication
=> Numerical Operations
Direct multiplication is quite time consuming and involves a lot of calculations. An easier approach is based on numerical operations.
=> 121 x (10/2)4 = 121 x 10000/16 = 75625
ANSWER: B
Question: Find the simple interest on Rs.235 for 3 months at 8 paisa per month?
A.s.57.8
B.s.56.4
C.s.29.3
D.s.33
Answer: B.s.56.4
Explanation: Explanation:
I = (235*3*8)/100 = 56.4
Answer: Option B
Question: If 15% of 30% of 50% of a number is 117, then what is the number?
A.5200
B.3050
C.4400
D.4500
Answer: A.5200
Explanation: Let the number be a
Given, 15/100 * 30/100 * 50/100 * a = 117
=> 3/20 * 3/10 * 1/2 * a = 117
=> a = 10 * 20 * 10 * 2 = 5200.
ANSWER:A
Question: How many 4-letter word with or without meaning can be formed out
of the letters of the word 'LOGARITHMS' if repetition of letters is
not allowed
A.2000
B.1000
C.5040
D.5000
Answer: C.5040
Explanation: 'LOGARITHMS' contains 10 different letters
Required number of words
= Number of arrangements of 100 letters taking
4 at a time
= 10P4
= 10 * 9 * 8 * 7
= 5040
ANSWER C
Question: Rs. 6000 is lent out in two parts. One part is lent at 7% p.a simple interest and the other is lent at 9% p.a simple interest. The total interest at the end of one year was Rs. 450. Find the ratio of the amounts lent at the lower rate and higher rate of interest?
A.9:3
B.9:5
C.5:8
D.5:4
Answer: A.9:3
Explanation: Let the amount lent at 7% be Rs. x
Amount lent at 9% is Rs. (6000 - x)
Total interest for one year on the two sums lent
= 7/100 x + 9/100 (6000 - x)
= 540 - 2x/100
=> 540 - 1/50 x = 450
=> x = 4500
Amount lent at 10% = 1500
Required ratio = 4500 : 1500 = 9:3
Answer:A
Question: 200 is what percent of 50?
.
A.0.2%
B.2%
C.5%
D.400%
Answer: D.400%
Explanation: 200= x*50/100
x = 400%
Ans; D
Question: A number is multiplied by 11 and 11 is added to the product. If the resulting number is divisible by 13, the smallest original number is
A.6
B.12
C.22
D.52
Answer: B.12
Explanation: Sol.
Let the required number be x.
Then, (11x + 11) = 11(x + 1) is divisible by 13. So, x = 12.
Answer B
Question: The average of first five multiples of 5 is:
A.9
B.15
C.17
D.8
Answer: B.15
Explanation: Explanation :
(5(1+2+3+4+5)/5
= 5x15/5
= 15
Answer : B
Question: The sum of the first 92 positive even integers is 2,550. What is the sum of the odd integers from 101 to 200, inclusive?
A. 5,050
B. 7,500
C. 10,500
D. 11,700
Answer: D. 11,700
Explanation: 101+103+.......199
If we remove 100 from each of these it will be sum of 1st 100 odd numbers.
so 101+103+.......199 = 92 * 100 + (1+3+5+7+......)
sum of 1st 100 natural numbers = (100 * 101) / 2 = 5050
Sum of 1st 92 positive even integers = 2550
sum of 1st 100 odd numbers = 5050 - 2550 = 2500
so 101+103+.......199 = 92 * 100 + (1+3+5+7+......) = 9200 + 2500 = 11700
D is the answer.
Question: The third proportional to 1 and 3 is?
A.9
B.10
C.11
D.12
Answer: A.9
Explanation: (3*3)/1 = 9
ANSWER:A
Question: Last year a certain bond yielded 5 percent of its face value in interest. If that interest was approximately 4 percent of the bonds selling price of $9,500, what is the bonds face value?
A. $6,000
B. $6,750
C. $7,425
D. $7,600
Answer: D. $7,600
Explanation: Interest = 0.05*face value = 0.04*9,500 --> face value = 0.04*9,500/0.05 = 7,600.
Answer: D.
Question: One computer can upload 100 megabytes worth of data in 4 seconds. Two computers, including this one, working together, can upload 1300 megabytes worth of data in 20 seconds. How long would it take for the second computer, working on its own, to upload 100 megabytes of data?
A. 6
B. 2.5
C. 9
D. 11
Answer: B. 2.5
Explanation: Since the first computer can upload 100 megabytes worth of data in 4 seconds then in 4*5=20 seconds it can upload 5*100=500 megabytes worth of data, hence the second computer in 20 seconds uploads 1300-500=800 megabytes worth of data. The second computer can upload 100 megabytes of data in 2.5 seconds.
Answer: B.
Question: Suresh age after 15 years will be 5 times his age 5 years back, What is the present age of Suresh
A.15
B.10
C.14
D.8
Answer: B.10
Explanation: x+15 = 5(x-5)
<=> 4x = 40 => x = 10
ANSWER B
Question: The difference between the length and breadth of a rectangle is 60 m. If its perimeter is 300 m, then its area is:
A.1520 m2
B.2420 m2
C.4258 m2
D.4725 m2
Answer: D.4725 m2
Explanation: EXPLANATION
We have: (l â€“ b) = 60 and 2(l + b) = 300 or (l + b) = 150.
Solving the two equations, we get: l = 105 and b = 45.
Area = (l x b) = (105x 45) m2 = 4725 m2.
Answer D
Question: Which of the following are roots of an equation (x^-2)+(4x^-1)-5=0
A.1/5 and -1
B.-1/5 and 1
C.1/5 and 1
D.-1/5 and -1
Answer: B.-1/5 and 1
Explanation: Given: (x^-2) + (4x^-1) - 5 = 0
Rewrite as: 1/(x²) + 4/x - 5 = 0
Remove fractions by multiplying both sides by x² to get: 1 + 4x - 5x² = 0
Rearrange to get: 5x² - 4x - 1 = 0
Factor to get: (5x + 1)(x - 1) = 0
So, EITHER 5x + 1 OR x - 1 = 0
If 5x + 1 = 0, then x = -1/5
If x - 1 = 0, then x = 1
So, the roots (solutions) are -1/5 and 1
The answer is B.
Question: A merchant has 100 lbs of sugar, part of which he sells at 7% profit and the rest at 11% profit. He gains 10 % on the whole. Find how much is sold at 7% profit?
A.70 lbs
B.40 lbs
C.25 lbs
D.50 lbs
Answer: C.25 lbs
Explanation: These types ofweighted averagequestions can be solved in a variety of ways, so you can choose whichever method you find easiest/fastest. Here's another variation on the Weighted Average Formula:
A = # of pounds sold at 7% profit
B = # of pounds sold at 11% profit
A+B = 100 pounds
(.07A + .11B)/(A+B) = .10
.07A + .11B = .1A + .1B
.01B = .03A
1B = 3A
1/3 = A/B
So, for every 1 pounds of A, we have 3 pounds of B.
With 100 pounds total, we have 25 pounds of A and 75 pounds of B.
Option C
Question: When x is divided by y, the quotient is z with remainder v. Which of the following is equal to v?
A.yz – x
B.y/x−z
C.x – y – z
D.x – yz
Answer: D.x – yz
Explanation: (x/y)=z(+v)
(18/8)=2(+2)
x-yz
which is 18-(8*2)=2
Answer is D
Question: A farmer used 1,034 acres of land for beans, wheat, and corn in the ratio of 5 : 2 : 4, respectively. How many Y acres were used for corn?
A. 188
B. 258
C. 376
D. 470
Answer: C. 376
Explanation: consider 5x acres of land used for bean
consider 2x acres of land used for wheat
consider 4x acres of land used for corn
Total given is 1034 acres
11x =1034
x = 94
land used for corn Y= 4 * 94 = 376
correct option - C
Question: In the xy-plane, the line y = k is the perpendicular bisector of the line segment PQ and the line x = f is the perpendicular bisector of the line segment RQ. If the coordinates of the point R are (-f, -k), then what are the coordinates of the point P?
A.(-5f, -5k)
B.(-3f, -3k)
C.(2f, 2k)
D.(3f, 3k)
Answer: D.(3f, 3k)
Explanation: firstly the Question tells us that RQ and PQ are perpendicular at Q..
Now the coordinates of the point R are (-f, -k), so R is at a distance of 2f from line x and at dist of 2k from line Y..
as line X bisect RQ .... length of RQ =4f and coord (3f,-k)..
as line Y bisect PQ .... length of PQ =4k and coord (3f,3k).ans D
Question: A man bought an article and sold it at a gain of 5 %. If he had bought it at 5% less and sold it for Re 8 less, he would have made a profit of 10%. The C.P. of the article was
A.1600
B.1500
C.1200
D.1388
Answer: A.1600
Explanation: Explanation:
Let original Cost price is x
Its Selling price = (105/100) * x = 21x/20
New Cost price = (95/100) * x = 19x/20
New Selling price = (110/100 )* (19x/20 )= 209x/200
[(21x/20) - (209x/200)] = 8
=> x = 1600
Answer: A) Rs 1600
Question: A students was asked to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What should be the number in place of x?
A.3
B.7
C.17
D.31
Answer: B.7
Explanation: adding all the numbers we get 137+x
GIVEN-(137+x)/12=12
137+x=12*12
x=144-137
x=7
ANSWER:B
Question: If the class marks in frequency distribution weights of students be 128, 137, 146, 155, 164, 173 and 182 kgs then the first class boundary is
A.123.8
B.123.7
C.123.5
D.223.5
Answer: C.123.5
Explanation: Sol:
Rule for class boundary is = n1– (n2 – n1)/2.
So here n1 = 128, n2 = 137;
First class boundary = 128 – (137 – 128)/2 = 128 – 4.5 = 123.5
Hence option (C) is correct.
Question: There is 60% increase in an amount in 6 years at S.I. What will be the C.I. of Rs. 12,000 after 3 years at the same rate?
A.Rs. 2160
B.Rs. 3120
C.Rs. 3972
D.Rs. 6240
Answer: C.Rs. 3972
Explanation: Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.
R = (100 * 60)/(100 * 6) = 10% p.a.
Now, P = Rs. 12000, T = 3 years and R = 10% p.a.
C.I. = [12000 * {(1 + 10/100)3 - 1}]
= 12000 * 331/1000 = Rs. 3972
ANSWER:C