Quantitative Aptitude

Time and Distance

Question 1: Two cities X and Y are 45 km apart. Two cars start from X and Y in the same direction with speeds of 30 km/hr and 15 km/hr, respectively. Both cars meet at point Z beyond Y. Find the distance ZY.

Solution: 
Relative speed = 30 – 15 = 15 km/hr
Time required by the faster car to overtake the slower car = Distance / Speed = 45 / 15 hr
Distance between Y and Z = 15 * (45 / 15) = 45 km

Question 2: Alice and Bob start from the same place M at the same time towards N, which are 60 km apart. Alice’s speed is 4 km/hr more than that of Bob. Alice turns back after reaching N and meets Bob at a 12 km distance from N. Find the speed of Bob.

Solution: 
Let the speed of the Bob = x km/hr 
Then Alice speed will be = (x + 4) km/hr 
Total distance covered by Alice = 60 + 12 = 72 km 
Total distance covered by Bob = 60 – 12 = 48 km 
Acc. to question, their run-time is same. so
72/ (x + 4) = 48/ x 
72x = 48x + 192 
24x= 192 
x= 8 
Bob speed is 8 km/hr.

You can improve your understanding of these topics by practicing more related questions. Access practice questions by clicking here.

Time and Work

Question 1: Person A can complete a task in 4 hours. When persons B and C work together, they can finish the task in 3 hours, and when persons A and C work together, they can complete it in 2 hours. What is the time it takes for person B alone to complete the task?

Solution: 
Person A 1 hour work = 1/4
Person (B + C) 1 hour work = 1/3
Person (A + C) 1 hour work = 1/2
Total work done by person A + B + C => (1/4 + 1/3) = 7/12
1 hour work of person B => 7/12 - 1/2 = 1/12
Time take by person B = 12 hours

Question 2: A, B, and C can individually complete a task in 10, 12, and 15 days, respectively. They start working together, but A leaves after 2 days, and B leaves 3 days before the completion of the task. Find the total number of days required to complete the entire task.

Solution:
 Total work done is LCM(10, 12, 15)=60 unit 
A’s efficiency = 60/10= 6 
B’s efficiency = 60/12= 5 
C’s efficiency = 60/15= 4 
First two days all work together 
So, the work completed in first two days= 15 x 2 = 30 unit 
Remaining work= 60 – 30 = 30 unit 
If B completes 3 day work also = 3 x 5 = 15 unit 
Total work remaining= 30 + 15 = 45 unit 
Number of days B and C works= 45/9=5 

You can improve your understanding of these topics by practicing more related questions. Access practice questions by clicking here.

Boats and Streams

Question 1: A person can row at a speed of 5 km/hr in still water. With a current velocity of 1 km/hr, it takes him 1 hour to row to a place and come back. What is the distance to the place?

Solution: 
Speed downstream = (5 + 1) kmph = 6 kmph.
Speed upstream = (5 - 1) kmph = 4 kmph.
Let the required distance be x km.
then, x/6 + x/4 = 1
10x/24 = 1
x = 2.4 km

Question 2: A boatman can row a boat at a speed of 5 km/h upstream and 15 km/h downstream. It takes him 2.5 hours to cover upstream and 10 hours to cover downstream. Find the speed of the stream and the speed of the boat in still water.

Solution: 
We are given that the boatman covers 5 km upstream in 2.5 hours and 15 km downstream in 10 hours. 
=> Speed upstream, U = 5 / 2.5 = 2 km / hr 
=> Speed downstream, D = 15 / 1.5 = 10 km / hr 
Therefore, Speed of boat in still water = 0.5 x (D + U) km / hr = 0.5 x (10 + 2) = 6 km / hr 
Also, speed of the stream = 0.5 x (D – U) km / hr = 0.5 x (10 – 2) = 4 km / hr

You can improve your understanding of these topics by practicing more related questions. Access practice questions by clicking here.

Pipes and Cisterns

Question 1: A tap drips at a rate of one drop per second, and it takes 800 drops to make 100 ml. The number of liters of water wasted in 30 days of a month is?

Solution: 
1 sec = 1 drop 
Number of seconds in 30 days= 30days x 24hrs x 60min x 60sec 
Number of milli-liters wasted = (100 x 30days x 24hrs x 60min x 60sec)/800 
= 324000 ml 
Number of liter= 324000/1000 
= 324 liters water wasted. 

Question 2: Two pipes, A and B, can individually fill a tank in 15 minutes and 20 minutes, respectively. When both pipes are opened simultaneously, they work together for 4 minutes. After this initial 4 minutes, pipe A is turned off. What is the total time needed to completely fill the tank?

Solution: 
Part filled in 4 minutes = 4 (1/15 + 1/20) = 7/15
Remaining part = (1 - 7/15) = 8/15
Part filled by B in 1 minute = 1/20
that is -> 1/20 : 8/15  ::  1 : x
x = (8/15 x 1 x 20) = 10 min 40s
so, total time needed is  = 14 min 40s

You can improve your understanding of these topics by practicing more related questions. Access practice questions by clicking here.

Profit and Loss

Question 1: A dealer wants to mark the price of an article such that by offering a 5 % discount, he is able to get 33 % profit. Find the percent of CP above which the article should be marked. 

Solution: 
Let the cost price of the article be Rs. 100 => Selling price of the article = Rs. 100 + 33% of CP = Rs. 133 
Let the marked price be Rs. M => Selling price = Marked Price – Discount =>
 133 = M – 0.05 M => 133 = 0.95 M => M = 140 => M – CP = 140 – 100 = 40 
Therefore, percent of CP above which the article should be marked = (40 / 100) x 100 = 40 %

Question 2: A man purchased an article of Rs 2500 and sold it at 40% above the cost price. If he has to pay Rs 250 as tax on it his net profit percentage will be:

Solution: 
P = 40%
Profit = CPx25/100
=> 2500×40/100
=> 1000
Net profit = profit – tax
=> 1000 – 250 = 750
Net profit % = 750/2500 x 100
=> 30%
Hence, net profit percent is 30%.

You can improve your understanding of these topics by practicing more related questions. Access practice questions by clicking here.

Simple Interest

Question 1: A person pays Rs 8000 as an amount on the sum of Rs 6000 that he had borrowed for 3 years. What will be the rate of interest?

Solution: 
A = Rs 8000
P = Rs 6000
Amount = Principal  + Simple Interest
SI = A – P 
    = 8000 – 6000 
    = Rs 2000
Time (t) = 3 years
Rate (R) = ?
SI = (P × R ×T) / 100
R = (SI ×100) /(P× T)
R = (2000 × 100 /(6000 × 3)  
   = 11.11 %
Thus, the rate of interest R is 11.11 %

Question 2: A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was:

Solution:
Principal  = Rs (100 x 5400 / 12 x 3) => Rs. 15000

You can improve your understanding of these topics by practicing more related questions. Access practice questions by clicking here.

Compund Interest

Question 1: A financier borrows funds at an annual interest rate of 5%, settling the interest at the close of the year. Subsequently, the financier extends loans at an 8% per annum compound interest rate, compounded semi-annually, and receives the accumulated interest at the conclusion of the year. Through this strategy, the financier achieves a gain of Rs 120.08 annually. The amount of money he borrows:

Solution:
Scenario (1): Interest Rate = 5% Time = 1 year
Scenario (2): With interest compounded semi-annually
the rate becomes 8/2 = 4% 
Time = 1 x 2 = 2 years 
Effective interest rate over 2 years = 4 + 4 + 4×4/100 = 8.16% Difference in rates = 8.16% – 5 = 3.16%
Given the gain is Rs 120.08, we establish the following relationship: 3.16 -> 120.08 1 -> 38 100 -> 3800
Therefore, the sum borrowed by the financier is Rs 3800.

Question 2: Given that the simple interest accrued on a certain amount over a period of 2 years at an annual rate of 5% is Rs. 50, determine the compound interest for the same principal, at the same rate, and over the same time frame.

Solution: 
Case (1): 
Simple Interest: Rate = 5% Time = 2 years 
Given Simple Interest = Rs. 50
Principal amount = 50 x 100 / 5 x 2
Principal = 500
Case (2):
Here we have: P=500 (principal amount) R=5% (rate) T=2 years (time)
Compund Interest = 500 x ((1 + 5/100)2-1)
on solving we get, Compound Interest = 51.25

You can improve your understanding of these topics by practicing more related questions. Access practice questions by clicking here.

Permutation and Combination

Question 1: What is the total number of possible motor vehicle registration plates for Haryana, each following the format (HR 12Q 8702), where the digits 0 to 6 are used without repetition, and a consonant is placed at the alphabetical position?

Solution: 
For Haryana state number plate always contains HR in starting. 
Two consonants already used in HR so remaining consonant = 21 -2 = 19. 
So, possible number of plates = 1(HR)x7x6x19(consonants)x5x4x3x2 = 95760

Question 2: How many different arrangements are possible for the letters in the word ‘LEADER’?

Solution: 
The word 'LEADER' has 6 letters 1L, 2E, 1A, 1D and 1R.
Required number of ways =6!/(1!)(2!)(1!)(1!)(1!) = 360

You can improve your understanding of these topics by practicing more related questions. Access practice questions by clicking here.

Probability

Question 1: If a pen is randomly drawn and replaced from a bag containing pens numbered 1 to 17, and this process is repeated, what is the probability that the first drawn pen has an even number, and the second drawn pen has an odd number?

Solution:
For the first drawn, we have 8 even numbered pens out of 15 and in second we have 9 odd numbered pens.
so, required probability = 8/17 x 9/17 = 72/289

Question 2: When three unbiased coins are flipped, what is the probability of getting either zero, one, or two heads?

Solution:
Here we have possibilities of all outcomes= {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most 2 heads
Then outcomes will be = {TTH, TTT, THT, HTT, THH, HTH, HHT}
p (E) = n (E) / n (S) = 7/8

You can improve your understanding of these topics by practicing more related questions. Access practice questions by clicking here.

Clock

Question 1: In a day, how many times it occur when the hands of a clock align in a straight line but move in opposite directions?

Solution:
The hands of a clock align in opposite directions (in a straight line) 11 times within a 12-hour period. 
This occurs twice every hour, except between 5 and 7 when it only happens once at 6 o'clock. 
Therefore, in a single day, the hands point in opposite directions a total of 22 times.

Question 2: Between 5 and 6 o’clock, at what specific time do the minute and hour hands form an angle of 34 degrees with each other?

Solution:
At 5 o'clock, the angle between the minute hand and the hour hand is 150 degrees. 
When this angle changes by either subtracting 34 degrees (150 - 34) 
or adding 34 degrees (150 + 34), it takes 21 1/11 minutes and 33 5/11 minutes, respectively.
Given that the angle changes by 5.5 degrees per minute,
the angle of 34 degrees occurs at two instances: 
at 5 hours 21 1/11 minutes and again at 5 hours 33 5/11 minutes.

You can improve your understanding of these topics by practicing more related questions. Access practice questions by clicking here.

Age

Question 1: The current age of a mother is three years greater than three times her daughter’s age. In three years, the mother’s age will exceed twice the daughter’s age by 10 years. Determine the mother’s current age.

Solution:
Let the daughter’s present age be ‘x’ years.
=> Mother’s present age = (3x + 3) years
So, according to the question
(3x + 3 + 3) = 2 (x + 3) + 10
=> 3x + 6 = 2x + 16
=> x = 10
Hence, mother’s present age = (3x + 3) = ((3 x 10) + 3) years = 33 years

Question 2: The current ages of Sameer and Anand are in the ratio 5:4. In three years, this ratio is expected to change to 11:9. What is Anand’s current age?

Solution:
Let the present ages of Sameer and Anand be 5x years and 4x years respectively.
Now, let's solve the equation:
(5x + 3) / (4x + 3) = 11/9
=> 9(5x + 3) = 11 (4x + 3)
=> 45x + 27 = 44x + 33
=> x = 6
Anand's current age is: 4x = 4 (6) = 24 years

You can improve your understanding of these topics by practicing more related questions. Access practice questions by clicking here.

Numbers

Question 1: What is the smallest digit that can replace * in the number 517*324 to ensure that the resulting number is divisible by 3?

Solution:
Sum of all digit must be divisible by 3. That means:
=> (5 + 1 + 7 + * + 3 + 2 + 4) = (22 + x), this must be divisible by 3.
x = 2;
2 will be come in place of * in the above number.

Question 2: When (67⁶⁷ + 67) is divided by 68, what will be the reminder?

Solution:
 (xn + 1) will be divisible by (x + 1) only and only when n is odd.
That means:
(6767 + 67) will be divisible by (67 + 1)
means:  (6767 + 1) + 66, when divided by 68 gives remainder of 66.
answer is 66.

You can improve your understanding of these topics by practicing more related questions. Access practice questions by clicking here.

Average

Question 1: The batsman’s average increased by 5 runs after scoring 120 runs in his 16th innings. Determine the batsman’s current average.

Solution:
Let the average of 15 innings is x
According to the question 
15x + 120 = 16(x + 5) 
15x + 120 = 16x + 80 
x = 40 
Hence, current average of the batsman is (40 + 5) = 45 

Question 2: The car owner purchases petrol at Rs. 7.50, Rs. 8, and Rs. 8.50 per liter in 3 consecutive years, spending Rs. 4000 each year. What is the approximate average cost per liter of petrol over the three years?

Solution:
Total quantity of petrol consumed in 3 years = (4000/ 7.50 + 4000/ 8 + 4000/ 8.50) liters
=> 4000 (2/15 + 1/8 + 2/17)
=> 76700/ 51 liters
Total amount spend = Rs. (3 x 4000) = 12000
Average cost = 12000 x (51/76700) = 6120/767 = Rs. 7.98

You can improve your understanding of these topics by practicing more related questions. Access practice questions by clicking here.

Percentage

Question 1: A student multiplied a number by 3/5 instead of 5/3. What will be the percentage error in the calculation?

Solution:
Let the number be x.
Then error will be, (5x/3 - 3x/5) = 16x/x5
Error % will be, (16x/15 x 3/5x x 100)% = 64%

Question 2: The broker applies a commission of 5% on orders up to Rs. 10,000 and 4% on orders exceeding Rs. 10,000. After deducting his commission, he remits Rs. 31,100 to his client. Determine the amount of the order.

Solution:
Let the order amount be Rs. n => Commission charged = >
 => 5 % of Rs. 10,000 + 4 % of (Rs. n – 10,000) 
=> Rs. 500 + 0.04 n – 400 
=> Commission charged = Rs. 100 + 0.04 n 
Now, amount remitted = Rs. n – (100 + 0.04 n) = 31,100 
=> 0.96 n – 100 = 31,100 
=> 0.96 n = 31200 
=> n = 32500 
Therefore, order amount = Rs. 32,500

You can improve your understanding of these topics by practicing more related questions. Access practice questions by clicking here.

Ratio and Proportion

Question 1: If x varies inversely as y-1 and is equal to 30 when y=8, find x when y=11.

Solution: 
x= k/ (y-1)
30 = k/(8-1)
k = 210
Put k=210 and y=11
x = 210/10 = 21

Question 2: In a 60-liter mixture, the ratio of milk to water is initially 2:1. If the desired ratio is to be changed to 1:2, what quantity of water needs to be additionally added?

Solution:
Quantity of milk  = (60 x 2/3) = 40 liters
Quantity of water in it = 60 - 40 => 20 liters
New ratio will be: 1 : 2
Let quantity of water to be added more with x liters.
Then, mile : water = > (40 / 20 + x)
Now, (40/ 20 + x) = 1/2
20 + x  80
x = 60 
Quantity to be added is  60 liters.

You can improve your understanding of these topics by practicing more related questions. Access practice questions by clicking here.

If we wish to excel in aptitude tests, interviews, placement exams, or competitive exams, we must prepare and practice well. However, to cover the commonly asked aptitude questions, we need proper guidance and direction. To enhance our problem-solving skills and sharpen our minds, a set of topics has been created. It includes two questions along with solutions to give us an idea of topics and types of questions that are usually asked. This article presents every topic and the commonly asked questions under it. By practicing these questions, we can get ourselves ready for the upcoming exam, test, and interview.