RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: A car averages 55 mph for the first 4 hours of a trip and averages 70 mph for each additional hour. The average speed for the entire trip was 60 mph. How many hours long is the trip?
A.6
B.8
C.11
D.12
Answer: A.6
Explanation: In 4hr journey = 55*4 = 220mph
Let in the next additional hour journey = 70*x =70x mph
So, (220+70x)/(4+x)= 60
=> 220+70x = 240+60x
=> 10x = 20
=> x = 2h
So the trip is (4+x)=> 6 hours long
ANSWER:A
Question: GOOD is coded as 164 then BAD coded as 21.if ugly coded as 260 then DUMP?
A.200
B.210
C.216
D.225
Answer: C.216
Explanation: GOOD=(7+15+15+4)*4=164
BAD= (2+1+4)*3=21
UGLY= (21+7+12+25)*4=260
coding follows
(sum of position of alphabets)*(no. of letters in the given word)
so,
DUMP=(4+21+13+16)*4= 216
ANSWER:C
Question: We bought 85 hats at the store. Blue hats cost $6 and green hats cost $7. The total price was $600. How many green hats did we buy?
A.A)36
B.B)90
C.C)40
D.D)42
Answer: B.B)90
Explanation: Let B be the number of blue hats and let G be the number of green hats.
B+G=85.
B=85-G.
6B+7G=600.
6(85-G)+7G=600.
510-6G+7G=600.
G=600-510=90.
The answer is B.
Question: A part of certain sum of money is invested at 8% per annum and the rest at 12% per annum, if the interest earned in each case for the same period is equal, then ratio of the sums invested is?
A.4:6
B.3:2
C.4:3
D.4:1
Answer: B.3:2
Explanation: 12:8
= 3:2
Answer: B
Question: 12. What is the dividend. divisor 17, the quotient is 9 and the remainder is 6.
A.150
B.152
C.154
D.159
Answer: D.159
Explanation: D = d * Q + R
D = 17 * 9 + 6
D = 153 + 6
D = 159
ANSWER D
Question: If LCM of two number is 693, HCF of two numbers is 11 and one number is 99, then find other
A.34
B.77
C.12
D.45
Answer: B.77
Explanation: Explanation:
For any this type of question, remember
Product of two numbers = Product of their HCF and LCM
So Other number =
693×1199
= 77
Answer: Option B
Question: 1000 men have provisions for 15 days. If 200 more men join them, for how many days will the provisions last now?
A.10.5
B.12.5
C.13.5
D.11.5
Answer: B.12.5
Explanation: 1000*15 = 1200*x
x = 12.5
ANSWER B
Question: Viswak was born 3 years after his father's marriage. His mother is 8 years younger than his father but 20 years older than Viswak who is 10 years old. At what age did the father get married ?
A.23 years
B.25 years
C.26 years
D.24 years
Answer: B.25 years
Explanation: Explanation:
Viswak's present age = 10 years.
His mother's present age = (10 + 20) years = 30 years.
His father's present age = (30 + 8) years = 38 years.
His father's age at the time of Viswak's birth = (38 - 10) years = 28 years.
Therefore Viswak's father's age at the time of marriage = (28 - 3) years = 25 years.
Answer: B
Question: If a person walks at 15 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance traveled by him is?
A.50 km
B.76 km
C.18 km
D.60 km
Answer: D.60 km
Explanation: Let the actual distance traveled be x km. Then,
x/10 = (x + 20)/15
5x - 300 =>x = 60 km.
Answer: D
Question: If x and y are both odd prime numbers and x < y, how many distinct positive integer Q factors does 2xy have?
A.3
B.4
C.6
D.8
Answer: D.8
Explanation: Since 2xy prime Q factors are x^1*y^1*2^1, its total number or factors must be (1+1)(1+1)(1+1)=2^3=8. Thus, I think D would be the correct answer.
Question: A grocer purchased 80 kg of sugar at 13.50 per kg and mixed it with 120 kg sugar at 16 per kg. At what rate should he sell the mixture to gain 17%?
A.17 per kg
B.17.55 per kg
C.16.5 per kg
D.16 per kg
Answer: B.17.55 per kg
Explanation: C.P. of 200 kg of mixture = (80 × 13.50 + 120 × 16)
= 3000.
S.P. = 117% of 3000 = (117⁄100 × 3000) = 3510
∴ Rate of S.P. of the mixture = (3510⁄200) per kg
= 17.55 per kg.
Answer B
Question: A train of 24 carriages, each of 60 meters length, when an engine also of 60 meters length is running at a speed of 60 kmph. In what time will the train cross a bridge 1.5 km long?
A.1 mins
B.3 mins
C.8 mins
D.6 mins
Answer: B.3 mins
Explanation: D = 25 * 60 + 1500 = 3000 m
T = 3000/60 * 18/5 = 180 sec
= 3 mins
Answer: B
Question: Find out the wrong number in the given sequence of numbers.
1, 2, 6, 15, 31, 57
A.2
B.57
C.15
D.31
Answer: B.57
Explanation: 1(result) + (1*1) = 2.
2(result) + (2*2) = 6.
6(result) + (3*3) = 15.
15(result) + (4*4) = 31.
31(result) + (5*5) = 56.
Now we are getting 56 not 57.. So 57 is the wrong number of the given.
answer : B
Question: If C(n, 7) = C(n, 5), find n
A.15
B.12
C.18
D.-1
Answer: B.12
Explanation: We know that C(n,r) = n! / r!(n - r)!
Now C(n,7) = C(n,5)
n! / 7! (n - 7)! = n! / 5! (n - 5)!
=> 5! (n - 5)! = 7! (n - 7)!
=> [5!] x [(n - 5)(n - 6)[(n - 7)!] = 7 x 6 x 5! x (n - 7)!
=> n2 - 11n + 30 = 42
=> n2 - 11n - 12 = 0
=> (n - 12)(n + 1) = 0
=> n = 12 or n = - 1
But n = - 1 is rejected as n is a non negative integer. Therefore, n = 12
ANSWER:B
Question: A sum of money at simple interest amounts to Rs. 2240 in 2 years and to Rs. 2600 in 5 years. What is the principal amount
A.1000
B.1500
C.2000
D.2500
Answer: C.2000
Explanation: Explanation:
SI for 3 year = 2600-2240 = 360
SI for 2 year 360/3 * 2 = 240
principal = 2240 - 240 = 2000
Answer: Option C
Question: Look at this series: 31, 29, 24, 22, 17, ... What number should come next?
A.11
B.15
C.16
D.17
Answer: B.15
Explanation: B
15
This is a simple alternating subtraction series, which subtracts 2, then 5.
Question: A Man travelled a distance of 80 km in 10 hours. He travelled partly on foot at 5 km/hr and partly on bicycle at 9 km/hr. What is the distance travelled on foot?
A.16 km
B.14 km
C.12 km
D.10 km
Answer: D.10 km
Explanation: Explanation:
Let the time in which he travelled on foot = x hour
Time for travelling on bicycle = (9 - x) hr
Distance = Speed * Time, and Total distance = 80 km
So,
5x + 10(9-x) = 80
=> 5x = 10
=> x = 2
So distance traveled on foot = 5(2) = 10 km
Answer: Option D
Question: At a conference, the only events scheduled for the day are seminars and breaks. All seminars are the same length and are 10 times as long as each break. A break must be between two consecutive seminars. If the first seminar begins at 9:00 a.m. and the third one ends at 11:20 a.m., how long will the conference last if there are a total of 6 seminars that day?
A.hours and 15 minutes
B.hours and 20 minutes
C.hours and 25 minutes
D.hours and 30 minutes
Answer: B.hours and 20 minutes
Explanation: Well its like this S B S B S B S B S B S
the interval between 9 am to 11:40 am is 160 min having 3 seminars and 2 breaks .If 1 break is of 5 min than time for 3 seminars = 160 - 2*5=150
Time of 1 seminar =150 /3=50 min .
So calculating for 6 seminars and 5 breaks ,total time comes out to be 5 hrs and 20 min .
Answer is B.
Question: When n is divided by 19, the remainder is 7. When n is divided by 11, the quotient is equal to Z, and the remainder is equal to 7. Z must be divisible by which of the following numbers?
A.16
B.18
C.20
D.19
Answer: D.19
Explanation: N = 19 q + 7
N = 11 z + 76
Equating
19 Q + 7 = 11z+7
11z=19q
11z/19 = q
since q is an integer Z must be divisible by 19.
Hence D
Question: In the formula A = (2r)^3, if r is halved, then A is multiplied by ?
A.64
B.1/8
C.1
D.8
Answer: B.1/8
Explanation: Say r=2 => A1 = 64
when r=1; V2 = 8
V2 = 1/8*V1.
Answer : B
Question: In a 8 x 8 chess board what is the total number of squares.
A.238
B.204
C.678
D.169
Answer: B.204
Explanation: Ans: The total number of squares in a n x n chess board is equal to "the sum of first n natural number squares"
i.e., n(n+1)(2n+1)6n(n+1)(2n+1)6
So Substituting 8 in the above formula we get 204
Answer:B
Question: Two circular signs are to be painted. If the radius of the larger sign is three times that of the smaller sign, how many times more paint is needed to paint the larger sign? (We can assume that a given amount of paint covers the same area on both signs.)
A.3
B.9
C.12
D.15
Answer: B.9
Explanation: Let R be the radius of the smaller sign.
The area A of the smaller sign is A = piR^2.
The area of the larger sign is pi(3R)^2=9piR^2=9A.
Since the area is 9 times larger, we need 9 times more paint for the larger sign.
The answer is B.
Question: If 20 machine can finish a job in 36 days, then how many more machines would be needed to finish the job in one -third less time?
A.a. 4
B.b. 8
C.c. 10
D.d. 12
Answer: C.c. 10
Explanation: You might think of this in a management context - we can use the principle of 'person-hours' to solve any problem where we have identical workers. So, using simpler numbers, suppose you know that 6 identical employees, working simultaneously, would finish a job in 5 hours. Then that job requires 6*5 = 30 total hours of person-work. If instead you wanted the job done in 3 hours, you'd assign 30/3 = 10 employees to do the job, because you want to get a total of 30 hours of work from the employees.
We can solve this problem identically. If 20 machines (identical ones, I assume) work simultaneously for 36 days, they will do a total of 20*36 machine-days of work. So the job requires 20*36 days of machine work in total. We instead want the job done in 1/3 less time, so in 24 days. So we'll need 20*36/24 = 30 machines, or 10 additional machines.
C
Question: In terms of percentage profit, which among following the best transaction.
A.C.P. 36, Profit 17
B.C.P. 50, Profit 24
C.C.P. 40, Profit 19
D.C.P. 60, Profit 29
Answer: D.C.P. 60, Profit 29
Explanation: Explanation:
Hint: Calculate profit percent as
Profit% = (profit/cost) * 100
ANSWER IS D
Question: A and B can do a piece of work in 3 days, B and C in 4 days, C and A in 6 days. How long will C take to do it?
A.18 days
B.20 days
C.24 days
D.30 days
Answer: C.24 days
Explanation: 2c = ¼ + 1/6 – 1/3 = 1/12
c = 1/24 => 24 days
ANSWER:C
Question: A, B, C and D enter into partnership. A subscribes 1/3 of the capital B 1/4, C 1/5 and D the rest. How much share did A get in a profit of Rs.2460?
A.14028
B.14029
C.14027
D.14000
Answer: D.14000
Explanation: 25*12: 30*12: 35*8
15:18:14
14/47 * 47000 = 14000
Answer: D
Question: Two cars A and B are running towards each other from different places 88 km apart. If the ratio of the speeds of the cars A and B is 5 : 6 and the speed of the car B is 90 km per hour then after how long will the two meet each other?
A.2263 minutes
B.24 minutes
C.32 minutes
D.36 minutes
Answer: C.32 minutes
Explanation: Speed of the car A = 5⁄6 × 90 = 75 km/hr
\ Reqd time = 88/90+75×60=32minutes
Answer C
Question: If A can do a work in 8 days and B can do the same work in days,in how many days A and B together can finish the work
A.3 3/7
B.4 3/7
C.5 3/7
D.6 3/7
Answer: A.3 3/7
Explanation: no of days taken to finish the work=8*6/(8+6)
Instead of solving this directly take the LCM of 8 and 6
8*6/(8+6)=LCM(8,6)/(LCM(8,6)/8 + LCM(8,6)/6)=24/(3+4)=24/7=3 3/7
Answer A
Question: The ratio of length of two refill is 9:8.2nd refill got damaged when only half length of 2nd refill get used.The ratio of total no. of pages written by both refill.
A.8:3
B.6:5
C.9:8
D.9:4
Answer: D.9:4
Explanation: the no. of page written is directly proportional to length of refill.
so, initial no. of page to be write =9x,8x.
now,2nd refill only can write 4x page
so new ratio of page written =9x/4x=9:4.
answer D
Question: if {x+y}/{x-y}= 4/3 and x≠0, then what percentage (to the nearest integer) of x+3y is x–3y ?
A.40%
B.30%
C.55%
D.36%
Answer: A.40%
Explanation: Explanation :
=>(x+y)/(x-y)=4/3.
=>(x/y)+1/(x/y)−1=4/3.
Cross-multiplying this equation yields
=>3x/y+3=4x/y−4.
Solving for x/y yields x/y=7.
Now, the percentage of x+3y the expression x−3y makes is
=>x−3y /x+3y∗100
Dividing both the numerator and the denominator of the expression by y, gives:-
=>((x/y)−3/(x/y)+3))×100.
=>((7−3)/(7+3))×100.
=>40%.
Answer : A
Question: Two trains leave the same train station at 6:00 am and 6:30 am and they travel at 100 kph and 120 kph respectively. How many kilometers from the train station will the two trains be together?
A.300
B.320
C.340
D.360
Answer: A.300
Explanation: At 6:30, the first train is 50 km ahead.
The second train gains 20 km each hour.
The time it takes the second train to catch the first train is 50 /20 = 2.5 hours.
In this time, the second train travels 2.5*120=300 km.
The answer is A.
Question: One hour after Matthew started waking from e to y, a distance of 45km, Johnny started walking along the same road from y to e. Matthew's walking rate was 3 km per hour and Johnny's was 4km per hour, how many km had Johnny walked when they met?
A.24
B.23
C.22
D.21
Answer: A.24
Explanation: Just an alternative method...
After the first hour the distance e is 42 km (45-3). Now the problem can be treated as if bothof them started at the same time. Since the speeds are in the ratio 3 : 4, the distances will also be in the same ratio. Splitting 42 in that ratio we get 18 : 24. So answer is 24.
Hence A.
Question: what is the Max number of rectangular boxes, each measuring 4 inches by 6 inches by 10 inches, that can be packed into a rectangular packing box measuring 16 inches by 18 inches by 30 inches, if all boxes are aligned in the same direction?
A.36
B.14
C.12
D.15
Answer: A.36
Explanation: The 4 inch side should be aligned to the 16 inch side ( 4 layer)
6 inch side should be aligned to the 18 inch side. (3 layer)
and 10 inch side should be aligned to the 30 inch side.(3 layer)
Maximum number of rectangles =4*3*3=36
Answer is A
Question: In an arithmetic progression the difference between the any two consecutive terms is a constant. What is the arithmetic mean of all of the terms from the first to the 29th in an arithmetic progression if the sum of the 13th and 17th terms of the sequence is 102?
A.43
B.47
C.51
D.55
Answer: C.51
Explanation: Let x be the difference between any two consecutive terms.
The mean of a sequence like this is the middle term, thus the 15th term in the sequence.
Then the mean of the 14th and 16th term is also equal to the overall mean, because the 14th term is (15th term - x) and the 16th term is (15th term + x).
Similarly the mean of the 13th and 17th term is also equal to the mean.
etc...
Thus the mean is 102/2 = 51
The answer is C.
Question: Hariprasad and Madhusudan started a business, investing sums in the ratio of 2 : 3. If Hariprasad had invested an additional amount of 10,000 the ratio of Hariprasad's investment to Madhusudan's investment would have been 3 : 2. What was the amount invested by Hariprasad?
A.8000
B.12000
C.9000
D.Data inadequate
Answer: A.8000
Explanation: Let the initial investments of Hariprasad and Madhusudan be 2x and 3x,respectively.
From the question,
2x+10000/3x=32
or, 4x + 20000 = 9x
\ x = 4000
\ Amount invested by Hariprasad = 2x = 8000
Answer A
Question: In a hot dog eating contest the average number of hot dogs that were eaten by boys was 7 each, and the average number eaten by girls was 5 each. If the total number of hot dogs eaten during the contest was 50, what was the total number of boys and girls in the contest?
A.6
B.7
C.8
D.9
Answer: C.8
Explanation: We can determine quickly that total number should range between 50/7<= N <= 50/5 , so ans should be between 7 and 10.
Now solving the expression 7A+5B =50
decreasing 50 by multiples of 7 and checking divisibility of that number by 5, we get
5 boys and 3 girls will total 50 hot dogs, 5*7 + 3*5 = 50
Hence total number is 5 + 3 = 8
Ans C
Question: A certain restaurant offers 8 different salads, 6 different main courses, 6 different desserts. If customers choose one salad, one main course and two different desserts for their meal, how many different meals are possible?
A.120
B.240
C.480
D.600
Answer: A.120
Explanation: MAGOOSHOFFICIAL SOLUTION:
To count the number of meals, we have to count the possibilities for the three components, and then, according to theFundamental Counting Principle, we multiply. For more on the FCP, seethis blog.
For salads, there are 8 possibilities. Done.
For main courses, there are 5 different possibilities. Done.
For desserts, there are 6 choices, but we pick two different choices. This would be 6C2. To calculate this, we can usethe shortcut, nC2 = (n*(n-1))/2.
6C2 = (6*5)/2 = 15
So, there are 15 combinations of two different desserts.
For the number of meals, we multiply 8*6*15=720
Answer = A
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 650 sq. feet, how many feet of fencing will be required?
A.34
B.40
C.85
D.88
Answer: C.85
Explanation: EXPLANATION
We have: l = 20 ft and lb = 650 sq. ft.
So, b = 32.5 ft.
Length of fencing = (l + 2b) = (20 + 65) ft = 85 ft.
Answer C
Question: If an article is sold at 19% profit instead of 12% profit, then the profit would be Rs. 105 more. What is the cost price?
A.1533
B.1543
C.1500
D.1278
Answer: C.1500
Explanation: Explanation:
Let the cost price of an article be Rs. x.
(19% of x) - (12% of x) = 105
19x/100 - 12x/100 = 105 => 7x = 105 * 100
=> x = 1500
Cost price = Rs. 1500
Answer:C
Question: A shipment of 1500 heads of cabbage, each of which was approximately the same size, was purchased for $600. The day the shipment arrived 1/3 of the heads were sold, each at 25% above the cost per head. The following day the rest were sold at a price equal to 5% less than the price each head sold for on the day before. What was the total gross profit on this shipment.
A.$100
B.b $115
C.c $120
D.d $125
Answer: D.d $125
Explanation: Purchase cost = 600/1500= $ 0.4
25% above the cost per head = (125/100)*0.4= $ 0.5
2/3 of the heads were sold means 1/3*1500 =500 ,Selling amount for 500 cabbage = 500 * 0.5 = $ 250
Remaining were sold at a price equal to 10% less than the price each head sold for on the day before i.e , (95/100)*0.5 = $ 0.475
Hence selling cost for the remaining cabbage = 1000 * 0.475 = $ 475
Total selling cost = 475 + 250 = $ 725
Gross profit = $ 725 - $ 600 = $ 125
Hence Answer is D
Question: 30 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work ?
A.20
B.77
C.36
D.25
Answer: A.20
Explanation: Explanation:
Less Men, means more Days {Indirect Proportion}
Let the number of days be x
then,
27 : 30 :: 18 : x
[Please pay attention, we have written 27 : 30 rather than 30 : 27, in indirect proportion, if you get it then chain rule is clear to you :)]
{\color{Blue} x=\frac{30\times 18}{27}}
x = 20
So 20 days will be required to get work done by 27 men.
Answer: A
Question: QUANT 4-PACK SERIES Problem Solving Pack 4 Question 2 For every positive integer n...
For every positive integer n that is greater than 1, the function a(n) is defined to be the sum of all of the odd integers from 1 to n, inclusive. The a(n) could have any of the following units digits except…?
A.1
B.2
C.4
D.6
Answer: B.2
Explanation: For every positive integer n that is greater than 1, the function a(n) is defined to be the sum of all of the odd integers from 1 to n, inclusive. The a(n) could have any of the following units digits except…?
Used brute force on this one to calculate. Not sure if there is a quicker way.
1+3=4 C is out
1+3+....15= 81 A is out
1+3+5+7=16 D is out
1+3+5=9 E i out
Answer: B
Question: A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 11 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?
A.13
B.13.5
C.14
D.14.5
Answer: B.13.5
Explanation: Ans is B
Given d_ab = 2*d_bc
let d_ab = d and d_bc = x so d=2x
for average miles per gallon = (d+x)/((d/12)+(x/11)) = 15.3 (formula avg speed = total distance/ total time)
Question: If 5 less than x is 2 greater than y, then a number z that is 5 greater than x reduced by y is equal to
A.12
B.9
C.14
D.y + 9
Answer: A.12
Explanation: 12 it is....nice one....
A
Question: A trader marked the selling price of an article at 10% above the cost price. At the time of selling, he allows certain discount and suffers a loss of 1%. He allowed a discount of :
A.10 %
B.10.5 %
C.11 %
D.12.5 %
Answer: A.10 %
Explanation: Sol.
Let C.P. = Rs. 100. Then, Marked Price = Rs. 110, S.P. = Rs. 99.
∴ Discount % = [11/110 * 100] % = 10%
Answer A
Question: Karen places a bet with Tom that she will beat Tom in a car race by 4 miles even if Karen starts 4 minutes late. Assuming that Karen drives at an average speed of 60 mph and Tom drives at an average speed of 45 mph, how many Y miles will Tom drive before Karen wins the bet?
A.15
B.18
C.21
D.24
Answer: D.24
Explanation: Let K and T be the speeds of Karen and Tom respectively.
t be the time that Karen will travel ----> t+4/60 will be the total time Tom will travel by the time the distance between Karen and Tom is 4 miles.
Thus, per the question , K(t) - T(t+4/60) = 4 ---> t = 7/15 hours
Thus the distance traveled by Tom when Karen is 4 miles ahead of him Y: T*(t+4/60) = 45 (7/15+4/60) = 24 miles. D is the correct answer.
Question: The price of a scooter and TV are in the ratio 7:5. If the scooter costs $100 more than the TV. then the price of the TV?
A.400
B.350
C.250
D.500
Answer: B.350
Explanation: Let the price of the scooter and TV is 7x and 5x
7x-5x = 100
x = 50
Price of TV = 7*50 = $350
Answer is B
Question: If four boxes of crayons contain six different colored crayons in each box, and a crayon is randomly selected from each box, what is the probability of getting at least one pair of the same colored crayons?
A.13/18
B.5/18
C.1/2
D.2/3
Answer: A.13/18
Explanation: AT LEAST one pair of matching crayons.
First solve for NO matching crayons:
1st box = any color = 6/6
2nd box = not a match to the first = 5/6
3rd box = not a match to the 1st or 2nd = 4/6
4th box = not a match to the 1st or 2nd or 3rd = 3/6
(6/6)(5/6)(4/6)(3/6) =
(1)(5/6)(2/3)(1/2) =
10/36
10/36 is the probability of NO matching crayons, so...
1 - 10/36 = 26/36 = the probability of getting at least one matching pair of crayons.
26/36 = 13/18
Answer: A
Question: A certain quantity is measured on two different scales, the R scale and the S scale, that are related linearly. Measurements on the R scale of 6 and 24 correspond to the measurements on the S scale of 30 and 60 respectively. What measurement on the R scale corresponds to a measurement of 75 on the S scale?
A.30
B.33
C.36
D.39
Answer: B.33
Explanation: A change of +18 on the R scale is a change of +30 on the S scale.
R = (18/30)*S + K = 3S/5 + K
6 = 3(30)/5 + K
K = -12
R = 3S/5 - 12
R = 3(75)/5 - 12 = 45-12 = 33
The answer is B.
Question: John invested $ 30000 at 7% per annum, He got 34347 after some period, The interest was calculated compoundly. Find the period?
A.2 Years
B.2 1/2 Years
C.3 Years
D.4 Years
Answer: A.2 Years
Explanation: P(1+(R/100))power of n
so
=30000(1+(7/100))power of n = 34347
=(107/100)Power of n=34347/30000
To simplify 34347/30000
We get 11449/10000
This is in the form of (107/100) power of 2
so The value of n is 2
Answer is 2 years
Question: The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:
A.35 years
B.40 years
C.50 years
D.45 years
Answer: B.40 years
Explanation: Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years.
Sum of the present ages of wife and child = (20 x 2 + 5 x 2) years = 50 years.
Husband's present age = (90 - 50) years = 40 years.
Answer: Option B
Question: A call center has two teams. Each member of Team A was able to process 2/5 calls as compared to each member of Team B. If Team A has 5/8 as many number of call center agents as Team B, what fraction of the total calls was processed by Team B?
A.3/2
B.3/4
C.4/5
D.1/2
Answer: C.4/5
Explanation: Let team B has 8 agents, so team A has 5 agents
Let each agent of team B picked up 5 calls, so total calls by team B = 40
So, each agent in Team A picked up 2 calls, so total calls for team A = 10
Fraction for team B = 40/(40+10) = 4/5 = Answer = C