RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: Let a, b, c be distinct digits. Consider a two-digit number ‘ab’ and a three-digit number ‘ccb’, both defined under the usual decimal number system, if (ab)2= ccb > 300, then the value of b is
A.1
B.0
C.5
D.6
Answer: A.1
Explanation: Explanation :
Given , (ab)2 = ccb.
[ Only 1 , 5 and 6 on squaring will result with same digits in units place. ]
Taking a =2, b =1,
=> (21)^2 = 441 > 300.
Hence, the required value of b = 1
Answer : A
Question: Can you find the missing number in the sequence given below?
10 17 24 11 18 ? 12 19 26 13 20 27
A.25
B.23
C.20
D.18
Answer: A.25
Explanation: Let's break the given series as below:
10 17 24
11 18 ?
12 19 26
13 20 27
Now read the number from left hand side from top to bottom as :
So the number that will replace '?' is 25
Answer : A
Question: A fort of 2000 soldiers has provisions for 50 days. After 10 days some of them left and the food was now enough for the same period of 50 days as before. How many of them left?
A.400
B.288
C.266
D.121
Answer: A.400
Explanation: 2000 ---- 50
2000 ---- 40
x ----- 50
x*50 = 2000*40
x=1600
2000
-------
400
Answer: A
Question: In how many different ways can 4 physics, 2 math, and 4 chemistry books be arranged in a row so that all books of the same subject are together?
A.5842
B.6912
C.7782
D.8132
Answer: B.6912
Explanation: Let's think of each subject as a group so there are 3 groups.
The number of ways to arrange these 3 groups is 3!
Within each subject the books can be re-arranged.
So we need to multiply 3! by 4!*2!*4! which is 3!*4!*2!*4! = 6912
The answer is B.
Question: How many seconds will a 500 metre long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr?
A.25
B.30
C.40
D.45
Answer: B.30
Explanation: Speed of the train relative to man	= (63 - 3) km/hr
= 60 km/hr
=(60 x(5/18))	m/sec
=50/3 m/sec.
Therefore Time taken to pass the man
=(500 x(3/50))sec
= 30 sec.
Answer :B
Question: A tin is full of wine. Out of this tin, 4 gallons of wine are taken out and 4 gallons of water are added. After this 4 gallons of mixture were taken out and again 4 gallons of water were added. Now the ratio of wine and water in the tin becomes 36:13. Find the volume of tin.
A.25
B.26
C.27
D.28
Answer: D.28
Explanation: We know wine/water = 36/13, therefore wine to total is 36/(36+13) or 36/49.
Out of the Total amount has been taken 4 gallons of wine, so the ratio of wine to total is (T-4)/T. The wine has been taken out twice, therefore, we put the ratio on a power of two and we make the following equation.
36/49 = ((T - 4)/T)^2
36 and 49 are both powers, so we can square the whole equation reaching the single equation of 6/7 = (T - 4)/T, solving it T=28.
ANSWER:D
Question: The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is:
A.3500
B.4000
C.4050
D.5000
Answer: B.4000
Explanation: Let P, Q and R represent their respective monthly incomes. Then, we have:
P + Q = (5050 x 2) = 10100 .... (i)
Q + R = (6250 x 2) = 12500 .... (ii)
P + R = (5200 x 2) = 10400 .... (iii)
Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 .... (iv)
Subtracting (ii) from (iv), we get P = 4000.
P's monthly income = Rs. 4000.
Answer: Option B
Question: A jar of 220 marbles is divided equally among a group of marble-players today. If 2 people joined the group in the future, each person would receive 1 marble less. How many people are there in the group today?
A.20
B.21
C.22
D.23
Answer: A.20
Explanation: 220 = 20*11 = 22*10
There are 20 people in the group today.
The answer is A.
Question: How long does a train 110 meters long running at the speed of 72 km/hour take to cross a bridge 132 meters in length ?
A.22.1 seconds
B.12.1 seconds
C.17.1 seconds
D.12.9 seconds
Answer: B.12.1 seconds
Explanation: Speed = 72 km/hour = 72*(5/18) m/sec
= 20 m/sec
Total distance to be covered = 110+132 = 142 meters
Time = Distance/Speed
= 242/20 = 12.1 seconds
Answer: B
Question: The time it took car P to travel 800 miles was 2 hours less than the time it took car R to travel the same distance. If car P’s average speed was 10 miles per hour greater than that of car R, what was car R’s average speed, in miles per hour?
A.43
B.58
C.60
D.70
Answer: B.58
Explanation: Let speed of car R be=x
Then speed of car P= x+10
A/Q,
(800/x)-(800/(x+10))=2
Solving for x=58 miles\hr.
B
Question: The total cars in a Garage is 1000 out of which 500 are made by Honda. Out of every 100 Honda cars, 90 are red in color, but only 50% of the total cars is red. Find the % of non-Honda cars that are red
A.10%
B.20%
C.30%
D.25%
Answer: A.10%
Explanation: Total Honda brand red cars = 90/100 *500 = 450
Total red cars = 50/100*1000 = 500
No. of Non-Honda cars = 1000-500 = 500
No. of Non-Honda red cars = 500 - 450 = 50
% of Non-Honda red cars = 50/500*100 = 10%
Answer : A
Question: By selling 150 mangoes, a fruit-seller gains the selling price of 30 mangoes. Find the gain percent?
A.15%
B.25%
C.115%
D.75%
Answer: B.25%
Explanation: Explanation:
SP = CP + g
150 SP = 150 CP + 30 SP
120 SP = 150 CP
120 --- 30 CP
100 --- ? => 25%
Answer: B
Question: How long will it take a sum of money invested at 5% p.a. S.I. to increase its value by 100%?
A.15 years
B.16 years
C.25 years
D.20 years
Answer: D.20 years
Explanation: Sol.
Let the sum be x. Then, S.I. = 100% of x = x; Rate = 5%.
âˆ´ Time = [100 * 2x/5 * 1/x*5] = 20 years.
Answer: D
Question: A worker's daily wage is increased by 50% and the new wage is $42 per day. What was the worker's daily wage before the increase?
A.$22
B.$24
C.$28
D.$32
Answer: C.$28
Explanation: Let x be the daily wage before the increase.
1.5x = $42
x = $28
The answer is C.
Question: In a partnership for a business, Jay invests Rs.6000 for complete year & Viru invests Rs.3000 for 6 months. What is Viru’s share if they earn Rs.240 as profit?
A.120
B.80
C.192
D.48
Answer: D.48
Explanation: Explanation:
If A invests amount C1 for T1 time and his share of profit is P1, and B invests amount C2 for T2 time and his share of profit is P2, then, C1 * T1 / C2 * T2 = P1/P2
If P is the Viru’s share of profit, then Jay gets (240 – P)
Therefore, 6000 * 12 / 3000 * 6 = (240 - P) / P = 72/ 18 = 4
4P = (240 – P)
5P = 240
P = 48
ANSWER D
Question: The smallest 6 digit number exactly divisible by 111 is?
A.100009
B.100010
C.100011
D.100015
Answer: C.100011
Explanation: The smallest 6-digit number 100000.
111) 100000 (900
999
-----
100
---
Required number = 100000 + (111 - 100)
= 100011
Option C
Question: Two same class students appeared at an examination. One of them secured 9 marks more than the other and her marks was 56% of the sum of their marks. What are the marks obtained by them?
A.45,30
B.42,33
C.50,40
D.55,45
Answer: B.42,33
Explanation: the marks secured by x => (x + 9)
Then sum of their = x + (x + 9) = 2x + 9
that (x + 9) was 56% of the sum of marks
=>(x+9) = 56/100(2x+9)
=>(x+9) = 14/25(2x+9)
=> 25x + 225 = 28x + 126
=> x = 33
=> (x + 9) = 33 + 9 = 42
->marks are 33 and 42
ANSWER B
Question: In what time will a railway train 60 m long moving at the rate of 36 kmph pass a telegraph post on its way?
A.7 sec
B.8 sec
C.2 sec
D.6 sec
Answer: D.6 sec
Explanation: T = 60/36 * 18/5
=6 sec
Answer:D
Question: What is the different between the place value and face value of 6 in the numeral 5468?
A.45
B.54
C.60
D.36
Answer: B.54
Explanation: place value of 6 = 6 * 10 = 60
face value of 6 = 6
60 - 6 = 54
Answer is B
Question: What is the difference between the place value of 2 in the numeral 7229?
A.150
B.160
C.180
D.190
Answer: C.180
Explanation: Option 'C'
200 - 20 = 180
Question: A certain list consists of 21 different numbers. If n is in the list and n is 8 times the average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?
A.1/20
B.1/6
C.2/7
D.4/21
Answer: C.2/7
Explanation: series : a1, a2....a20,n
sum of a1+a2+...+a20 = 20*x (x = average)
so, n = 8*x
Hence, a1+a2+..+a20+n = 28x
So, the fraction asked = 8x/28x = 2/7
C
Question: What is the area of a square field whose sides have a length of 16 meters?
A.225 sq m
B.286 sq m
C.298 sq m
D.256 sq m
Answer: D.256 sq m
Explanation: 16 * 16 = 256 sq m
The answer is D.
Question: The ratio of ages of A and B 1 year ago was 3:4.After 1 year it will be 5:6.what is the present age of A.
A.1
B.2
C.3
D.4
Answer: D.4
Explanation: Solution:Let x be the age of A and y be the age of B
hen ratio of their ages one year ago (x-1):(y-1)=3:4
=>(x-1)/(y-1)=3/4............(1)
Ratio of their ages after one year (x+1):(y+1)=5:6
=>(x+1)/(y+1)=5/6..........(2)
Solving (1) and (2)
(1)=>4(x-1)=3(y-1)
=>4x-4=3y-3
=>4x-3y=1.......................(3)
(2)=>6(x+1)=5(y+1)
=>6x-5y=-1......................(4)
3*(3) - 2*(4)
12x - 9y=3 -
12x - 10y=-2
___________
y = 5
___________
(3)=>4x-15=1
=>4x=16
=>x=16/4=4
Answer D
Question: In an examination, 20% of total students failed in Hindi, 70% failed in English and 10% in both. The percentage of these who passed in both the subjects is :
A.10%
B.20%
C.30%
D.40%
Answer: B.20%
Explanation: pass percentage = 100 - (20+70-10) = 100-80 = 20
ANSWER:B
Question: A can do a job in 9 days and B can do it in 27 days. A and B working together will finish twice the amount of work in ------- days?
A.22 days
B.18 days
C.22 6/2 days
D.27 days
Answer: B.18 days
Explanation: 1/9 + 1/27= 3/27 = 1/9
9/1 = 9*2 =18 days
Answer: B
Question: The speed at which a man can row a boat in still water is 15 kmph. If he rows downstream, where the speed of current is 3 kmph, what time will he take to cover 60 metres?
A.11
B.77
C.16
D.12
Answer: D.12
Explanation: Speed of the boat downstream = 15 + 3 = 18 kmph
= 18 * 5/18 = 5 m/s
Hence time taken to cover 60 m = 60/5 = 12 seconds.Answer: D
Question: Two goods trains each 500 m long are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one?
A.22
B.28
C.48
D.277
Answer: C.48
Explanation: Relative speed = 45 + 30 = 75 km/hr.
75 * 5/18 = 125/6 m/sec.
Distance covered = 500 + 500 = 1000 m.
Required time = 1000 * 6/125 = 48 sec.Answer:C
Question: If the length of a rectangle is halved and its breadth is tripled, what is the percentage change in its area?
A.25% increase
B.25% decrease
C.50% increase
D.50% decrease
Answer: D.50% decrease
Explanation: Explanation:
Length is halved.
i.e., length is decreased by 50%50%
Breadth is tripled
i.e., breadth is increased by 200%200%
Change in area
=(−50+200−50×200/100)%=50%
i.e., area is increased by 50%
Answer: Option D
Question: Frank is 12 years younger then John. In 5 years John will be twice as old as Frank. How old will Frank be in four years?
A.8
B.11
C.12
D.14
Answer: B.11
Explanation: Frank is 15 years younger then John --> F+12=J;
In 5 years John will be twice as old as Frank (in 5 year John will be J+5 years old and Frank will be F+5 years old) --> J+5=2*(F+5) --> (F+12)+5=2*(F+5) --> F=7;
In 4 years Frank will be 7+4=11 years old.
Answer: B.
Question: How many such 3's are there in the following number sequence which are immediately Preceded by an odd number and immediately followed by an even number ? 5 3 8 9 4 3 7 2 3 8 1 3 8 4 2 3 5 7 3 4 2 3 6
A.1
B.2
C.3 three
D.4
Answer: C.3 three
Explanation: A is followed by B means : A comes first,B comes next
A preceded by B means : B comes first, A comes next
The sets which satisfy the condition is: 5 3 8
1 3 8
7 3 4
ANSWER:C
Question: Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:
A.9
B.11
C.13
D.15
Answer: D.15
Explanation: Let the three integers be x, x + 2 and x+ 4.
Then,
3x = 2(x + 4) + 3
x = 11
Third integer = x + 4 = 15.
ANSWER:D
Question: By investing in 1623% stock at 64, one earns Rs. 1700. The investment made is
A.s. 9600
B.s. 7500
C.s. 5640
D.s. 6528
Answer: D.s. 6528
Explanation: Explanation :
Market Value = Rs.64
Face Value is not given and hence take it as Rs.100
16 2/3% of the face value = 50/3
ie, to earn 50/3, investment = Rs.64
Hence, to earn Rs.1700, investment needed = 64×3×1700/50= 6528
Answer : Option D
Question: At a circular table for eight will sit six adults and two small children. The two children will sit next to each other, and the two parents of the children will sit next to the children so that there is one parent on each side of the two children. If rotating their positions around the table is not considered changing the arrangement, in how many different ways can the eight people be arranged around the table.
A.24
B.96
C.1440
D.5040
Answer: B.96
Explanation: Lets consider P|C|C|P as one unit.
Number of people sitting in the circular table = 4 + 1unit = 5
Number of ways of arranging the above 5 units seated in the circular table = (5 - 1)! = 4! = 24
But children can interchange the positions between them in 2 ways and the parents can interchange their positions in 2 ways.
Total number of ways = 24 * 2 * 2 = 96
Answer: B
Question: A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
A.650
B.690
C.698
D.700
Answer: C.698
Explanation: S.I. for 1 year = Rs. (854 - 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (815 - 117) = Rs. 698 Answer: C
Question: Robert ate 12 chocolates, Nickel ate 3 chocolates. How many more chocolates did Robert ate than nickel?
A.A)4
B.B)7
C.C)9
D.D)5
Answer: C.C)9
Explanation: 12-3=9. Answer is C
Question: One-third of Rahul's savings in National Savings Certificate is equal to one-half of his savings in Public Provident Fund. If he has Rs. 1,00,000 as total savings, how much has he saved in Public Provident Fund ?
A.80000
B.40000
C.50000
D.75000
Answer: B.40000
Explanation: Let savings in N.S.C and P.P.F. be Rs. x and Rs. (100000 - x) respectively. Then,
=1/3 x = 1/2 (100000 - x)
=x/3 + x/2 = 50000
=5x/6 = 50000
=x = 50000 x 6 /5	= 60000
Savings in Public Provident Fund = Rs. (100000 - 60000) = Rs. 40000
Answer is B.
Question: A producer of tea blends two varieties of tea from two tea gardens one costing Rs 18 per kg and another Rs 20 per kg in the ratio 5 : 3. If he sells the blended variety at Rs 25 per kg, then his gain percent is
A.12%
B.33%
C.14%
D.15%
Answer: B.33%
Explanation: Explanation:
Suppose he bought 5 kg and 3 kg of tea.
Cost Price = Rs. (5 x 18 + 3 x 20) = Rs. 150.
Selling price = Rs. (8 x 25) = Rs. 200.
Profit = 200 - 150 = 50
So, Profit % = (50/150) * 100 = 33%
Option B
Question: A large tank can filled by A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half?
A.10min
B.15min
C.30min
D.25min
Answer: C.30min
Explanation: Part filled by A+B in 1 minute = 1/60 + 1/40 = 1/24
Suppose the tank is filled in x minutes
Then, x/2(1/24 + 1/40) = 1
x/2 * 1/15 = 1
x = 30 min
Answer is C
Question: The ratio of number of boys and girls in a class is 3 : 2. In the 1st semester exam 20% of boys and 25%
of girls get more than or equal to 90% marks. What percentage of students get less than 90% marks ?
A.56
B.70
C.78
D.80
Answer: C.78
Explanation: Expl : Let boys =3x and girls =2x.
Number of those who get less than 90% mark =(80% of 3x) + (75% of 2x)
=(80/100 )*3x +(75/100 *2x) =39x/10
Required percentage = (39x/10 *1/5x *100)% = 78%.
Answer:C
Question: If a person walks at 10 km/hr instead of 8 km/hr, he would have walked 14 km more. The actual distance traveled by him is?
A.50
B.40
C.56
D.16
Answer: C.56
Explanation: Let the actual distance traveled be x km. Then,
x/8= (x + 14)/10
2x - 112=>x = 56 km.
Answer:C
Question: The average temperature for Tuesday, Wednesday and Thursday was 52°C. The average temperature for Wednesday, Thursday and Friday was 54°C. If the temperature on Friday be 53°C, what was the Temperature on Tuesday?
A.39°C
B.44°C
C.37°C
D.47°C
Answer: D.47°C
Explanation: Explanation:
T + W + T = 52 × 3 = 156°C
W+ T+ F = 54 × 3 = 162°C
Also, temperature on Friday = 53°C
Temperature on Tuesday = 156 +53 - 162 = 47°C
Answer: Option D
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 r 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 - r)
B.x/(r - y)
C.x
D.x/r
Answer: B.x/(r - y)
Explanation: Two ways...
1) let t be the total number, so cost = t*y+x and selling price=t*r..
No profit no loss....t*y+x=t*r.....T=x/(r-y)...
B
Question: Let n~ be defined for all positive integers n as the remainder when (n - 1)! is divided by n.
What is the value of 34~ ?
A.6
B.1
C.2
D.0
Answer: D.0
Explanation: n~ = (n-1)!
so 34~ = (34-1)! = 33!
when 33!/34 we have 16*2 inside 33!
hence 34 gets cancelled and we get remainder as 0
D
Question: In a weight-lifting competition, the total weight of Joe's two lifts was 450 pounds. If twice the weight of his first lift was 300 pounds more than the weight of his second lift, what was the weight, in pounds, of his first lift?
A.250
B.275
C.325
D.350
Answer: A.250
Explanation: This problem is a general word translation. We first define variables and then set up equations.
We can define the following variables:
F = the weight of the first lift
S = the weight of the second lift
We are given that the total weight of Joe's two lifts was 450 pounds. We sum the two variables to obtain:
F + S = 450
We are also given that twice the weight of his first lift was 300 pounds more than the weight of his second lift. We express this as:
2F = 300 + S
2F – 300 = S
We can now plug in (2F – 300) for S into the first equation, so we have:
F + 2F – 300 = 450
3F = 750
F = 250
Answer is A.
Question: The ratio by weight, measured in pounds, of books to clothes to electronics in a suitcase initially stands at 5:4:2. Someone removes 9 pounds of clothing from the suitcase, thereby doubling the ratio of books to clothes. How many pounds do the electronics in the suitcase weigh?
A.7
B.8
C.9
D.10
Answer: C.9
Explanation: The weights of the items in the suitcase are 5k, 4k, and 2k.
If removing 9 pounds of clothes doubles the ratio of books to clothes, then 9 pounds represents half the weight of the clothes.
2k = 9 pounds and then k = 4.5 pounds.
The electronics weigh 2(4.5) = 9 pounds.
The answer is C.
Question: A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 10 sq. feet, how many feet of fencing will be required?
A.34
B.40
C.68
D.21
Answer: D.21
Explanation: We have: l = 20 ft and lb = 10sq. ft.
So, b = 0.5 ft.
Length of fencing = (l + 2b) = (20 + 1) ft = 21 ft.
ANSWER:D
Question: if x-y-z = 0,3 x+4y+3z = 4,5x+2y+7z = 22 ,what is the value of z?
A.3
B.4
C.5
D.2
Answer: A.3
Explanation: Explanation:
4x + 2y - 5z = - 21 â€“-------I
2x - 2y + z = 7â€“------ II
4x + 3y - z = - 1-------- III
Solve the first equation for x.
x-y-z = 0
x = y+z
Substitute the solution for x into the second and third equations.
II---> 3*(y+z)+4y+3z = 4
3y+3z+4y+3z = 4
7y+6z = 4
III---->5*(y+z)+2y+7z = 22
5y+5z+2y+7z = 22
7y+12z = 22
Subtract the new second equation from the new third equation and solve for z.
7y+12z = 22
- (7y+6z = 4)
6z = 18
z=3
Answer is A
Question: The output of a factory is increased by 10% to keep up with rising demand. To handle the holiday rush, this new output is increased by 20%. By approximately what percent would the output of the factory now have to be decreased in order to restore the original output?
A.20%
B.24%
C.30%
D.32%
Answer: B.24%
Explanation: Let the output of a factory is 100
First,increased by 10 %=110
due to holiday rush,it increase by 20%=110+22=132
132-(x/100*132)=100
24.24
ANSWER:B
Question: A boat with still water speed of a is traveling upstream where the streams speed is b. After covering certain distance s, the boat turns back and tries to come to the starting place. But the boat stops at 3/4th of the intended distance due to a technical fault. What is the ratio of the time taken for the return journey to that of the upstream journey.
A.3(a - b)/4(a + b)
B.4(a - b)/3(a + b)
C.3(a + b)/4(a - b)
D.4/3
Answer: A.3(a - b)/4(a + b)
Explanation: speed of boat upstream = a - b
speed of boat downstream = a + b
For upstream journey, time taken = distance / speed = s / (a - b)
During return journey, the boat had covered on 3/4th of s :
Therefore for return journey, time taken = distance / speed = (3s/4)/(a+b)
Ratio of time taken for return journey to the upstream journey = (3s/4)/(a+b)/ (s/(a - b)) = 3(a - b)/4(a + b)
ANSWER:A
Question: Two taps can separately fill a cistern 10 minutes and 15 minutes respectively and when the waste pipe is open, they can together fill it in 18 minutes. The waste pipe can empty the full cistern in?
A.5
B.6
C.7
D.9
Answer: D.9
Explanation: 1/10 + 1/15 - 1/x = 1/18
x = 9
Answer: D
Question: If p and q are the roots of the equation x2 - bx + c = 0, what is the equation if the roots are (pq + p + q) and (pq - p - q)?
A.x2 - 2cx + (c2 - b2) = 0
B.x2 - 2bx + (b2 + c2) = 0
C.bcx2 - 2(b + c)x + c2 = 0
D.x2 + 2bx - (c2 - b2) = 0
Answer: A.x2 - 2cx + (c2 - b2) = 0
Explanation: Explanatory Answer
In the given quadratic equation x2 - bx + c = 0,
The sum of the roots p + q = b --- (1)
And the product of the roots pq = c --- (2)
We have to formulate a quadratic equation whose roots are (pq + p + q) and (pq - p - q).
The sum of the two roots = pq + p +q + pq - p - q = 2pq
But from eqn (2), we know that pq = c
Therefore, the sum of the roots = 2c
The product of the roots = (pq + p + q)(pq - p - q)= (pq)2 - (p+q)2
From equation (1) and (2), we know that pq = c and p + q = b
Therefore, the product of the roots = c2 - b2
We know the sum of the roots and the product of the roots.
Therefore, the quadratic equation is x2 - (sum of the roots)x + product of the roots = 0
=> x2 - 2cx + c2 - b2 = 0
Answer A
Question: An amount of Rs. 100000 is invested in two types of shares. The first yields an interest of 9% p.a and the second, 11% p.a. If the total interest at the end of one year is 9 2/3 %, then the amount invested in each share was?
A.23777
B.35000
C.29977
D.26777
Answer: B.35000
Explanation: Let the sum invested at 9% be Rs. x and that invested at 11% be Rs. (100000 - x). Then,
(x * 9 * 1)/100 + [(100000 - x) * 11 * 1]/100 = (100000 * 29/3 * 1/100)
(9x + 1100000 - 11x)/100 = 29000/3 = 9700
(9x + 1100000 - 11x) = 970000
x = 65000
Sum invested at 9% = Rs. 65000
Sum invested at 11% = Rs. (100000 - 65000) = Rs. 35000.
Answer: B
Question: The number of bacteria in a petri dish increased by 50 percent every 2 hours. If there were 108 million bacteria in the dish at 2: 00 p.m., at what time were there 32 million bacteria in the dish?
A.6: 00 p.m.
B.8: 00 p.m.
C.6: 00 a.m.
D.8: 00 a.m.
Answer: D.8: 00 a.m.
Explanation: Given:
The number of bacteria in a petri dish increased by 50 percent every 2 hours
There were 108 million bacteria in the dish at 2: 00 p.m
Since the bacteria is increasing by 50% every 2 hours, this means that the number is multiplied by 1.5 every 2 hours.
Required: At what time were there 32 million bacteria in the dish?
Assume that there were n intervals of 2 hours between 32 million and 108 million.
32 * (1.5)^n = 108
(3/2)^n = 108/32 = 27/8 = (3/2)^3
Hence n = 3 intervals.
Each interval was of 2 hours. Therefore, the number of hours = 6
Subtracting 6 hours from 2 PM, we get 8 AM as the answer.
Option D
Question: If a and b are positive integers and x = 9^a and y = 5^b, which of the following is a possible units digit of xy?
A.1
B.4
C.5
D.7
Answer: C.5
Explanation: The units digit of 9^a is either 1 or 9.
The units digit of 5^b is 5.
1*5 = 5 and 9*5 = 45.
The units digit of xy is 5.
The answer is C.