Aptitude Tricks and Tips for Placement and Exams

What is Aptitude?

Aptitude means inherent competence to undertake specific tasks. It is the potential, an intrinsic property in an individual and helps undertake particular types of tasks. It is also related to the natural ability to learn particular things. For example, individuals with good aptitude in mathematics are highly likely to do well in computer science as this ability comes naturally to them.

What are the basics of aptitude?


An aptitude test shows the mental ability of a candidate in order to judge a candidate performance in different situations. It is a standard way to get maximum score. we are trying to aware you about basic concepts of aptitude such as numerical computation ability, analytical abilities.

How many aptitudes are there?

Most people have about four or five strong talents out of the roughly two dozen independent aptitudes known to exist. Most jobs require about four or five. As many as 10% of the population has double that number of aptitudes–and that is a problem for them and their employers.

What are some of the best aptitude tips and tricks?


There are no any hard and fast rules for aptitude.
Though some exist in case of axioms, but in general aptitude no such tricks exist. All that are there are derived from the actual methods only.
So, I would suggest you to better go for the problem solving technique because it will always have an upper hand than tricks. (since there are small tricks for different problems) thus it makes very difficult to remember all the tricks..
Some formulas are there, in profit and loss, percentage, Ci, si,permutations,binomial,probability and mensuration …just memorize them.

What are quantitative aptitude tricks and shortcuts for the SSC CGL tier 1?

Concepts are like stairs and tricks are like lifts.
Lift may fail but stairs wont. SSC CGL 2017 mains is an example. Question pattern and style changed and those whose concepts were clear, survived.
And tricks are nothing but a short way developed out of conceptual method. If you practice hard enough you will develop your own tricks and method.
Though the net is filled with plethora of tricks and tips page. But believe me I am saying this after giving 3 attempts at ssccgl and 4+ years of teaching experience behind me.
Focus on concept. In the moment of pressure when the clock ticks student always forget the tricks or mess up the trick method. However a trick practiced like a habit is engraved.
So practice as much you can. CGL aptitude pathshala ,dineshmiglani, e1 coaching center and abhinaymaths on youtube are good channels to follow online and explore for yourself.

Why won’t aptitude tricks work for badly remembered students?

Here remembering is not important, it just matter of understanding question properly.rest all is u a have to apply the knowledge u learnt till SSLC.
When you practice more questions you’ll gets more ideas to solve a problem.

Is any website for learning aptitude (short tricks)?

Those who are new to aptitude these are tricks on some important aptitude topics that might help you.
And for all the formulae of each topic you can follow books like R S agrawal. It has a great content.

What are some aptitude test solving tricks?

Important Formulas of Number System

Formulas of Number Series
1. 1 + 2 + 3 + 4 + 5 + … + n = n(n + 1)/2
2. (12 + 22 + 32 + ….. + n2) = n ( n + 1 ) (2n + 1) / 6
3. (13 + 23 + 33 + ….. + n3) = (n(n + 1)/ 2)2
4. Sum of first n odd numbers = n2
5. Sum of first n even numbers = n (n + 1)

Mathematical Formulas

1. (a + b)(a – b) = (a2 – b2)
2. (a + b)2 = (a2 + b2 + 2ab)
3. (a – b)2 = (a2 + b2 – 2ab)
4. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
5. (a3 + b3) = (a + b)(a2 – ab + b2)
6. (a3 – b3) = (a – b)(a2 + ab + b2)
7. (a3 + b3 + c3 – 3abc) = (a + b + c)(a2 + b2 + c2 – ab – bc – ac)
8. When a + b + c = 0, then a3+ b3 + c3 = 3abc
9. (a + b)n = an + (nC1)an-1b + (nC2)an-2b2 + … + (nCn-1)abn-1 + bn

Shortcuts for number divisibility check

1. A number is divisible by 2, if its unit’s digit is any of 0, 2, 4, 6, 8.
2. A number is divisible by 3, if the sum of its digits is divisible by 3.
3. A number is divisible by 4, if the number formed by the last two digits is divisible by 4.
4. A number is divisible by 5, if its unit’s digit is either 0 or 5.
5. A number is divisible by 6, if it is divisible by both 2 and 3.
6. A number is divisible by 8, if the number formed by the last three digits of the given number is divisible by 8.
7. A number is divisible by 9, if the sum of its digits is divisible by 9.
8. A number is divisible by 10, if it ends with 0.
9. A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places, is either 0 or a number divisible by 11.
10. A number is divisible by 12, if it is divisible by both 4 and 3.
11. A number is divisible by 14, if it is divisible by 2 as well as 7.
12. Two numbers are said to be co-primes if their H.C.F. is 1. To find if a number, say y is divisible by x, find m and n such that m * n = x and m and n are co-prime numbers. If y is divisible by both m and n then it is divisible by x.

Shortcuts for ‘recurring decimal to fraction’ conversion

1. For recurring decimals of format ‘0.abababab…’ (ab repeats), equivalent fraction will be “repeating group (here ab)”/”as many 9’s as the number of digits in repeating group”
2. For recurring decimals of format ‘0.abbbbb…’ (b repeats), equivalent fraction will be (entire decimal group – non-repeating decimal group)/(as many 9’s as the number of repeating digits in the decimal part with as many 0’s as the number of non-repeating digits in the decimal part)
Frequently Asked Questions on Number System
• Given a number x, you will be asked to find the largest n digit number divisible by x.
• You will be given with a set of numbers (n1, n2, n3…) and asked to find how many of those numbers are divisible by a specified number x.
• Given a number series, find the sum of n terms, find nth term etc.
• Find product of two numbers when their sum/difference and sum of their squares is given.
• Find the number when divisibility of its digits with certain numbers is given.
• Find the smallest n digit number divisible by x.
• Which of the given numbers are prime numbers.
• Number x when divided by y gives remainder r, what will be the remainder when x2 is divided by y.
• Given relationship between the digits of number, find the number.
• Find result of operations (additions, subtractions, multiplications, divisions etc) on given integers. These integers can be large and the question may look difficult and time consuming. But mostly the question will map onto one of the known algebraic equations given in this first tab.

What tips you know about cracking aptitude test?

A couple of weeks before I appeared for CAT, I attended a workshop by a famous coaching institute.
I had done well in some of their mock tests, so I got an invite to this workshop that promised to take your scores from high 90s to even higher 90s. Like an accelerator.
I went there with the hope that it’ll be the perfect prep right before the exam.
And boy, did it surprise me.
Out of 50 questions that we worked on, I could literally only solve 3. 3/50. 2 weeks before CAT.
I did what anyone in a tricky position would do. I absolutely ignored it and stuck to what was going well for me.
Fast forward a couple of months, and I had scored 99.82 percentile in CAT. A dream score for me, to say the very least.
So when I looked back at what went wrong in the workshop, I realized that the problem was the type of questions. The most complicated, formula-heavy questions I had every come across. If you didn’t remember a formula, there was nothing you could do about the question. Kind of like a test in school.
In other words, it was testing anything but your aptitude. And actual aptitude tests like CAT would never do that.
The key? Stick to stuff that challenges your thinking and not your memory. Improve your ability to think through logic.

If percentage of profit made, when an article is sold for Rs.78, is twice as when it is sold for Rs.69, the cost price of the article is

(A) Rs. 49
(B) Rs. 51
(C) Rs. 57
(D) Rs. 60

How do I crack any company’s aptitude test?

Aptitude exams are taken to filter out the students. They aren’t interested in testing your maths skills rather they are interested in testing in which direction you think. Aptitude exams are taken to test our ability to analyze the question, out thought process and the approach we choose to get to the answer.
What is required is basic knowledge of 8–9–10th maths. Further, practice the questions as much as possible. R.S.Aggarwal is one of the finest book, also Arun Sharma’s Quantitative Aptitude is equally good. So buy one of this book as soon as possible and start preparing from it.
The topics to be more focused are:
• Problems on Numbers
• HCF and LCM
• Average
• Problems on Ages
• Percentage
• Profit and Loss
• Ratio and Proportion
• Time and Work
• Pipes and Cisterns
• Time and Distance
• Problems on Trains
• Boats and Streams
• Simple and Compound Interest
• Area
• Calendar and Clocks
• Permutation and Combination
• Probability
• Height and Distance
• Tabulation Problems

What are the tricks to solving aptitude questions?

Learn the concept clearly and then practice, practice and practice.
There is no readymade shortcut. Once u practice, accuracy increases and time taken to solve the question decreases.
Also with the repeatation of questions, u automatically learn which step could be skipped or can be done in smarter way.

1. Distance =Average speed × time taken
2. Average speed = Distance travelled/time taken
3. Time taken = Distance travelled/Average Speed
4. X km/hour = x*5/18 meter per second
5. Y meter/second = y×18/5 kilometer per hour.

ABC is triangle If Sin (A+B/ 2) =√3/2, then the value of Sin C/2 is

(A) 1/√2
(B) 0
(C) 1/2
(D) √3/2

Solve the equation 2x + 5 = 9
A. 2
B. 5
C. 7
D. 9

By working 5 hours/day, 4 waiters can serve 250 dishes in 8 days. 2 waiters would require how many hours/day to serve 500 dishes in 20 days?

A. 8 hrs/day
B. 6 hrs/day
C. 7 hrs/day
D. 9 hrs/day

There are 25 horses among which you need to find out the fastest 3 horses. You can conduct race among at most 5 to find out their relative speed. At no point you can find out the actual speed of the horse in a race. Find out how many races are required to get the top 3 horses.

A. 5
B. 7
C. 8
D. 6

The daily income of two persons are in the ratio of 4:7. If each receives an increment of $10 ibn the daily income, the ratio is altered to 3:5, find theri daily salaries.
A. $120 & $210
B. $180 & $315
C. $80 and $ 140
D. $200 & $350

In a certain coding system ‘APRIL’ is written as ‘CSVNR’. How will you code ‘AUGUST’?

What are tips and tricks for cracking TCS aptitude test?

Short tricks play a very vital role in cracking any exam, especially aptitude exams. When I had my placements I was not much aware of these tricks, then a friend suggested that I must visit once and check their study materials.
I didn’t have enough pocket money to buy any such course, so I contacted and told them about my problem.
So they told me about the free sections on their website where I can find a good number of practice questions along with the tips and tricks and some help for solving questions easily.

Some important shortcuts you must know for cracking TCS aptitude test are:
• Butterfly method
• Chocolate method
• Successive method
• Skeleton method
• Dove method
• Steer and Ride

What are the main part of an aptitude test?

There are mainly three parts in an aptitude test-
A. Numerical reasoning
B. Logical ability
C. Verbal ability
But as per the needs of the job, there are other kinds of aptitude tests as well that are used for specific roles. For example, visual reasoning is used for the assessment of designers.

How can I improve my aptitude?

A. Practice reasoning test questions: To improve your problem solving skills it is essential that you practice answering the types of questions that are likely to appear in your official numeracy test. When you go through your practice materials do not guess the answer but rather think about the solution and try to arrive at the correct answer yourself. At first, study the challenge that you are struggling with slowly. Make sure that you understand the logic and important concepts to work out the problem.

B. Practice multiple-step solutions:In order to advance your aptitude further you need to have regular exposure to the types of questions that cannot by merely solved by applying single routine arithmetic operations to one step problems; but rather you need to be confronted with set of challenges that require you to tackle multiple-step solutions where you have to get through various stages of calculations or mental computations to get to the final answer.

C. Improve in your weak areas: To get ahead in your preparation you need to analyse any weaknesses in your knowledge of maths, mental arithmetic and problem solving. Any deficiencies in any of these areas may significantly impact on your ability to solve the questions and ultimately will lower your score. For example, if you have gaps in knowledge of sums and you have good mental or reasoning intellect then you are likely to fail because numbers cannot be manipulated except the rules of maths. Contrary, if you have good knowledge of arithmetic, that is, you know how to use percentages, fractions or other related arithmetic operations whereas your mental or reasoning abilities are not developed to the level that is required for numerical aptitude test, then your score is not likely to be high.

What is an example of aptitude?

Aptitude is your natural ability to learn or excel in a certain area. It is often the case that a person has a group of aptitudes that fit together and helps them succeed at specific tasks. For example, you could have an aptitude for math and logic.

Find the greatest 5 digit number divisible by 5, 15, 20, and 25
A. 99900
B. 99000
C. 99990
D. 90990

Solution: LCM of 5, 15, 20, and 25 is 300
The greatest 5 digit number is 99999
\frac{99999}{300}30099999 = 99
Therefore, required number 99999 – 99 = 99900

The product of two numbers is 3888. If the H.C.F. of these numbers is 36, then the greater number is:
A. 110
B. 108
C. 36
D. 120

Solution: Let the two numbers be 36x and 36y
Now, 36x * 36y = 3888
xy =\frac{3888}{36 × 36}36×363888
xy = 3
Now, co-primes with product 3 are (1, 3).
Therefore, the required numbers are 36 * 1 = 36
36 * 3 = 108
Therefore the greatest number is 108

Sum of two numbers is 60 and the H.C.F. and L.C.M. of these numbers are 5 and 100 respectively, then the sum of the reciprocals of the numbers is equals to:
A. \frac{3}{25}253
B. \frac{11}{220}22011
C. \frac{21}{120}12021
D. \frac{11}{320}32011

Solution: Let the numbers be a and b.
Now, given a+b = 60
a × b = HCF × LCM = 5 × 100
= 500
\frac{1}{a} +\frac{1}{b} = \frac{a+b}{a × b}a1+b1=a×ba+b
\frac{1}{a} +\frac{1}{b} = \frac{60}{500}a1+b1=50060

Suppose there are three different containers contain different quantities of a mixture of Sugar and rice whose measurements are 403 grams, 434 grams and 465 grams What biggest measure must be there to measure all the different quantities exactly?
A. 31 grams
B. 21 grams
C. 41 grams
D. 30 litres

Solution : Prime factorization of 403,434 and 465 is
H.C.F of 403, 434 and 465=31

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?
A. 8
B. 16
C. 9
D. 10

Solution : L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
Hence, the bells will toll together after every 120 seconds(2 minutes).
Therefore, in 30 minutes ,number of times bells toll together is \frac{30}{2} + 1230+1 = 16

Two people P and Q start running towards a circular track of length 400 m in opposite directions with initial speeds of 10 m/s and 40 m/s respectively. Whenever they meet, P’s speed doubles and Q’s speed halves. After what time from the start will they meet for the third time?

Solution : Time taken to meet for the 1st time= \frac{400}{40+10}40+10400=8 sec.
Now P’s speed = 20m/s and Q’s speed=20 m/s.
Time taken to meet for the 2nd time= \frac{400}{20+20}20+20400 = 10 sec.
Now P’s speed =40 m/sec and Q’s speed = 10 m/sec.
Time taken to meet for the 3rd time= \frac{400}{10+40}10+40400=8 sec.
Therefore, Total time= (8+10+8) = 26 seconds.

Find HCF of 12 and 16.
(A) 5
(B) 4
(C) 12
(D) 16

Solution Find the difference between 12 and 16. The difference is 4. Now, check whether the numbers are divisible by the difference. 12 is divisible by 4 and 16 is divisible by 4.Hence, the HCF is 4.

Find HCF of 18 and 22.
(A) 2
(B) 4
(C) 18
(D) 36

Solution: Find the difference between 18 and 22. The difference is 4. Now, check whether the numbers are divisible by the difference. Both 18 and 22 are not divisible by 4. So take the factors of the difference. The factors of 4 are 2*2*1. Now, check whether the numbers are divisible by the factors. 18 and 22 are divisible by factor 2.
Hence, the HCF is 2.

Find LCM of 2,4,8,16.
(A) 16
(B) 18
(C) 12
(D) 2

Solution Factorize of above number
2 =2
8 = 23
16 = 24

Choose the largest number. In this example, the largest number is 16. Check whether 16 is divisible by all other remaining numbers. 16 is divisible by 2, 4, 8. Hence, the LCM is 16.
Find the LCM of 2,3,7,21.
(A) 21
(B) 44
(C) 36
(D) 42

Solution: Choose the largest number. The largest number is 21. Check whether 21 is divisible by all other remaining numbers. 21 is divisible by 3 and 7 but not by 2. So multiply 21 and 2. The result is 42. Now, check whether 42 is divisible by 2, 3, 7. Yes, 42 is divisible. Hence, the LCM is 42.

The least prime number is
A 0

B 1

C 3

D 2

A natural number other than 1, is a prime number if and only if it is divisible by 1 and the number itself. Thus, the least prime number is 2.

Find the least number which divided by 12, 18, 36 and 45 leaves the remainder 8,14, 32 and 41 respectively.

Here, 12-8 = 18-4=36-32=45-41=4
The required number = (LCM of 12, 18,36 and 45)-4
2 = 12,18,36,45
2 = 6,9,18,45
2 = 3,9,9,45
3 = 1,3,3,15
3 = 1,1,1,5
5 = 1,1,1,1
LCM = 2*2*3*3*5=180
The required number = 180-4 = 176

An empty pool being filled with water at a constant rate takes 8 h to fill 3/5th of its capacity. How much more time will it finish filling the pool ?

Solution: Time taken to fill the 3/5th part of a pool = 8h
Remaining part = [ 1 – (3/5)]
= [(5-3)/5]=2/5th
⸫ Time taken to fill the 2/5th part of a pool = (8*2*5*)/(5*3) = 16/3
=5 1/3 = 5 h 20 min.

P can do a piece of work in 9 days. Q is 50% more efficient than P. The number of days it takes for Q to do the same piece of work is .

Solution: Let efficiency of P be 1.
⸫ Efficiency of Q=1+50% of the efficiency of P.
= 1 +(50/100)×1 = 1+(1/2)= 3/2
Since, P can do a piece of work in 9 days.
Then, Q can do a piece of work in 9*(2/3) = 6 days.

16 men can complete a work in 15 days, 24 children can do same work in 20 days. In how many days will 8 men and 8 children, complete the same work?

∵ 16 men can do a piece of work in 15 days.
⸫ 1 man can do a piece of work in 1 days = 1 /(15*16) = 1/240
1 child can do a piece of work in
1 day = 1/24*20 = 1/480
So, 240 men’s one day work = 480
Children’s one day work or 1 man’s one day work = 2
Children’s one day work
⸫ 8 men + 8 children
= (8*2+8) children
= (16+8) children = 24 children
Then, 24 children can do a piece of work in 480/24 i.e., 20 days.

A and B started a business in partnership with Rs.50,000 and Rs.60,000 respectively. A is also a working partner and gets 10% of the total profit for looking after the business. How much is the share or B less than the share of A in the profit of Rs.55,000 ?

Solution : Ratio of two Capital of A and B
= 50,000 : 60,000
= 5 : 6
∵∵ A gets 10% of the total profit for looking after the business.
∴∴ Income received by A for looking after the business.
= 10% Rs.55, 000
= Rs. 5500
Rest of the profit
= 55,000 – 5500
= Rs. 49500
Now, from this amounts of profit A and B will get their shares of profit in the ratio of their invested capitals.
Profit received by
A = 49,500×51149,500×5/11
= Rs.22,500
∴∴Total share of
A = 22500 + 5500
= Rs.28,000
Again share of
B = 49,500×61149,500×6/11
= 27000
∴∴ Hence share of B is Rs 1,000 = [28,000-27,000] less than the share of A.
Short Method :
∵∵ Since A gets 10% of the total profit for looking after the business therefore, B will get the share of profit from 90% of the total point.
∴∴ Share of profit B gets
= 55000×6/11×90/100
= Rs.27,000
∴∴ Share of income A gets
= 55,000 – 27,000
= Rs.28,000
∴∴ Required difference
= Rs. (28,000 – 27,000)
= 1000.

The partners A, B, C start a business. Twice the investment of A is equal to thrice the capital of B and the capital of B is four times the capital of C. Find the share of each out of a profit of Rs. 2,97, 000.

Solution : Let C’s capital = Rs. x then B’s capital = Rs. 4x
Now, 2 (A,s capital)= 3 (B’s capital)=3×4×3×4x
∴∴ A’s capital= (3×4×2)(3×4×2) = 6x
So, ratio of share of A, B and C = 6x : 4x : x = 6 : 4 : 1.
∴∴ A’s share =Rs.(297000×6/11)(297000×6/11)
= Rs.1,62,000
B’s share of =Rs.(297000×4/11)(297000×4/11)
= Rs.108000
and C’s share =Rs.(297000×1/11)(297000×1/11)
= Rs.27000.

A began a business with Rs. 2100 and is joined afterwards by B with Rs. 3600. After how many months did B join, if the profits at the end of the year are divided equally ?

Solution : Suppose B joines after x months. Then B’s money remained invested for (12-x) months.
∵2100×12∵2100×12= 3600×(12−x)3600×(12−x)
⇒⇒3600 x = 43200-25200
∴∴ x = 18000/3600 = 5
So, B, joined after 5 months.

Dilip and Manohar started a business by investing Rs. 1,00,000 and Rs. 1,50,000 respectively. Find the share of each out of profit of Rs. 24,000.
Solution : Ratio of share of Dilip and manohar
= 100000 : 150000 = 2 : 3
∴∴ Dilip’s share
= Rs.(24000×2/5)(24000×2/5)
= Rs.9600
Manohar’s share
= Rs.(24000×3/5)(24000×3/5)
= Rs.14400

Sanjay and Raju started a business and invested Rs.20,000 and Rs.25,000 respectively. After 4 months, Raju left Naresh jointed by investing Rs. 15,000. At the end of the year there was a profit of Rs. 4,600. What is the share of Naresh ?
Solution : Ratio of share of Sanjay, Raju and Naresh
= 20,000×12:25,000×4:15000×820,000×12:25,000×4:15000×8
= 12 : 5 : 6:
∴∴ share of Naresh
= Rs.(4600×6/23)(4600×6/23) = Rs.1200
If nc10=nc14nc10=nc14 find the value of n

Solution: nc10=nc14⇒n=(10+14)=24nc10=nc14⇒n=(10+14)=24

If nc3=220nc3=220 then find the value of n.

Solution: ∴n!(n−3)!3!=220∴n!(n−3)!3!=220
n(n-1)(n-2) = 1320 or n(n-1)(n-2) = 11×12×2111×12×21 or n= 12

If ncr+ncn+1=n+1cxncr+ncn+1=n+1cx then find the value of x.

Solution: ncr+ncn+1=n+1cr+1ncr+ncn+1=n+1cr+1, formula
ncr+ncn+1=n+1cxncr+ncn+1=n+1cx given
by comparing above two statement we can get , x = r+1

In how many ways can letter of the word ‘APPLE’ be arranged?

Solution: there is in all 5 letter .out of these two are p,one is A one is L and one is E.

A rectangular tin sheet is 12 cm long and 5 cm broad. It is rolled along its length to form a cylinder by making the opposite edges just to touch each other. Then, the volume of the cylinder is

Here, length of the rectangular in sheet = 12
And breadth of the rectangular tin sheet = 5 cm

When the rectangular sheet is rolled along its length into a cylinder so that the width of the sheet becomes the height of the cylinder, then the length of the sheet equals the circumference of the base of the cylinder.
i.e., 2πr = 12
r = 12/2π = 6/π cm and height, h = 5 cm
⸫ Volume of the cylinder = πr2h
= π(6/π)2 × 5 = π × (36/π2)×5
= 180/π cm3

A merchant advertise 10% off on the items bought from his store. The total discount got by a customer who bought a cooker worth 650, a heater worth 500 and a bag worth 65 is

Here, cost of a cooker = Rs. 650
Cost of heater = Rs. 500
And cost of a bag = Rs.65
Total cost = Rs.(650+500+65) = Rs. 1215
Discount = 10%
⸫ Net discount = Rs.(10% of 1215)
= Rs.[(10/100)*1215]
= 121.50

What is a better investment 4% stock at Rs.120 or 3% stock at Rs.80?

Solution: 4% stock at Rs. 120 = 4% of 120
= (4/100)*120 = Rs. 4.80 and 3% Stock at Rs. 80 = 4% of 80
= (4/100)*120 = 4% of 80
= Rs. 2.40
It is clear that first investment is better

20% raise of price followed by a discount of 25% of the raised portion will

Solution: Let original cost of the object be Rs. 100
If 20% raised in price, then cost of the object = Rs.120
Cost increased = Rs.(120-100) = Rs 20
⸫ Discount on increased cost
= 25% of 20 = (25/100)*20= 5
⸫ New cost = Rs. (120 – 5)
= Rs. 115.

The sum of the ages of two brothers, having a different of 8 yr between them, will double after 10 yr. What is the ratio of the age of the younger brother to that of the elder brother?

Solution: Let the present age of two brother’s be x yr and (X+8) yr.
After 10 yr, Age of two brothers (X+10)
And (X+8+10)
According to the question,
X+10+X+18 = (X+X+8)*2
→ 2X+28=(2X+8)*2
→ 2(X+14)=(2X+8)*2
→ X+14 = 2X+8
→ X-2X = 8-14
→ -X = -6
→ X = 6

The ratio of two numbers is 3:4 and their LCM is 180. The second number is

Solution :Let the two number be 3x and 4x.
According to the question
12x = 180
X= 180/12=15
Thus, First number = 3*15 = 45 and
second number = 4*15=60

A gun is fired at a distance of 1.34 km from Geeta. She hears the sound after 4 s. The speed at which sound travels is
Solution : Sound covers 1.34×1000 m distance in 4 s.
⸫ Speed of the sound = (1.34*1000)/4
= 1340/4=335 m/s

Without any hesitation, you can ask me anything in the comment section related quantitative aptitude shortcut tricks and methods. Visit on the next page for more practice or know more tricks and methods.

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