YSR NAVASAKAM

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Simplification 3

Simplification 3

Welcome to aptitude tricks
Hello Aspirants. Welcome to Online Quantitative Aptitude Section with explanation in AffairsCloud.com. Here we are creating question sample in Simplification. We have included Some questions that are repeatedly asked in bank exams !!!
  1. (4438 – 2874 – 559) ÷ (269 – 106 – 83) = ?
    A. 14.5
    B. 12.5
    C. 20.5
    D. 27.5
    E. None of these
    Answer & Explanation
    Answer – B. 12.5
    Explanation :
    4438 – 2874 – 559 = 1005 ; 269 – 106 – 83 = 80; 1005 ÷ 80 = 12.5
  2. (78.95)² – (43.35)² = ?
    A. 4353.88
    B. 4153.88
    C. 4253.78
    D. 4053.78
    E. None of these
    Answer & Explanation
    Answer – A. 4353.88
    Explanation :
    (78.95)² – (43.35)² = 6233.1025 – 1879.2225 = 4353.88
  3. 434.43 + 43.34 + 3.44 + 4 + 0.33 = ?
    A. 455.54
    B. 485.54
    C. 475.54
    D. 465.54
    E. None of these
    Answer & Explanation
    Answer – B. 485.54
    Explanation :
    434.43 + 43.34 + 3.44 + 4 + 0.33 = 485.54
  4. (755% of 523) ÷ 777 = ?
    A. 5
    B. 6
    C. 8
    D. 7
    E. None of these
    Answer & Explanation
    Answer – A. 5
    Explanation :
    (755/100 * 523) = 3948.65
    3948.65 ÷ 777 = 5
  5. 156 + 16 * 1.5 – 21 = ?
    A. 126
    B. 149
    C. 141
    D. 159
    E. None of these
    Answer & Explanation
    Answer – D. 159
    Explanation :
    16 * 1.5 = 24
    156 + 24 – 21 = 159
  6. 783.559 + 49.0937 * 31.679 – 58.591= ? 
    A. 1280
    B. 3280
    C. 2280
    D. 1880
    E. None of the above
    Answer & Explanation
    Answer –C. 2280
    Explanation :
    49.0937 * 31.679 = 1555
    783 + 1555 – 58 = 2280(approx)
  7. (√7921 – √2070.25) * (1/4) = ?
    A. 15
    B. 16
    C. 17
    D. 19
    E. 11
    Answer & Explanation
    Answer – E. 11
    Explanation :
    √7921 = 89 ; √2070.25 = 45.5 = 43.5/4 = 10.8 ≈ 11
  8. (12.25)– √625 = ?
    A. 145.1625
    B. 125.0625
    C. 155.1625
    D. 165.0625
    E. None of these
    Answer & Explanation
    Answer B. 125.0625
    Explanation :
    (12.25)2 = 150.0625
    150.0625 – 25 = 125.0625
  9. 8451 + 793 + 620 – ? = 6065 + 713
    A. 3486
    B. 3586
    C. 3286
    D. 3186
    E. None of these
    Answer & Explanation
    Answer – E. None of these
    Explanation :
    8451 + 793 + 620 = 9864 ;  6065 + 713 = 6778 => 9864 – 6778 = 3086
  10. 22240 ÷ √? = 34 * 12
    A. 3065
    B. 3085
    C. 3025
    D. 3075
    E. None of these
    Answer & Explanation
    Answer – C. 3025
    Explanation :
    34 * 12 = 408
    √? = 22240/408 = 54.5 = 55 => x = 3025
Simplification 2

Simplification 2

Hello Aspirants.
Welcome to Online Quantitative Aptitude Section with explanation in AffairsCloud.com. Here we are creating question sample in Simplification. We have included Some questions based on latest pattern !!!
  1. 69 ÷ 3 * 0.85+ 14.5 – 3 = ?
    A. 36.15
    B. 32.15
    C. 33.05
    D. 32.05
    E. None of these
    Answer & Explanation
    Answer – E. None of these
    Explanation :
    69 ÷ 3 * 0.85 = 23 * 0.85 = 19.55
    19.55 + 14.5 – 3 = 31.05
  2. 4.5 + 23.50 + 14.58 – 17.68 * 0.5 = ?
    A. 23.74
    B. 33.74
    C. 34.74
    D. 36.74
    E. None of these
    Answer & Explanation
    Answer – B. 33.74
    Explanation :
    17.68 * 0.5 = 8.84
    42.58 – 8.84 = 33.74
  3. 23.56 + 4142.25 + 134.44 = ?
    A. 4010.05
    B. 4000.15
    C. 4100.25
    D. 4300.25
    E. None of these
    Answer & Explanation
    Answer – D. 4300.25
    Explanation :
    23.56 + 4142.25 + 134.44 = 4300.25
  4. (√ 7744 * 66 ) ÷ (203 + 149)= ?
    A. 12.5
    B. 14.5
    C. 13.5
    D. 18.5
    E. 16.5
    Answer & Explanation
    Answer – E. 16.5
    Explanation :
    (√ 7744 * 66 )= 5808; 5808/352 = 16.5
  5. 5/9 of 504 + 3/8 of 640 =  ?
    A. 620
    B. 550
    C. 520
    D. 480
    E. 460
    Answer & Explanation
    Answer – C. 520
    Explanation :
    5/9 of 504 = 280 ; 3/8 of 640 = 240 => 280 + 240 = 520
  6. (786*74) ÷ x = 1211.75
    A. 62
    B. 55
    C. 52
    D. 48
    E. 46
    Answer & Explanation
    Answer – D. 48
    Explanation :
    (786*74) ÷ x = 1211.75
    x = 48
  7. 18² + √? = 350
    A. 676
    B. 576
    C. 26
    D. 28
    E. None of these
    Answer & Explanation
    Answer – A. 676
    Explanation :
    18² + √? = 350
    √? = 350 – 324 = 26
    ? = 676
  8. 140% of 500 + 24% of 750 = ?
    A. 550
    B. 660
    C. 770
    D. 880
    E. None of these
    Answer & Explanation
    Answer – D. 880
    Explanation :
    700 + 180 = 880
  9. 4900 ÷ 28 * 444 ÷ 12 = ?
    A. 6312
    B. 6223
    C. 6475
    D. 6217
    E. 6421
    Answer & Explanation
    Answer – C. 6475
    Explanation :
    4900 ÷ 28 = 175
    444 ÷ 12 = 37
    175 * 37 = 6475
  10. 1221 + 1117 = x% of 6680 
    A. 26
    B. 32
    C. 43
    D. 17
    E. 35
    Answer & Explanation
    Answer – E. 35
    Explanation :
    2338/66.8 = 35
Simplification 1

Simplification 1

Welcome to aptitude tricks
Hello Aspirants.
Welcome to Online Quantitative Aptitude Section with explanation in AffairsCloud.com. Here we are creating question sample in Simplification. We have included Some questions based on latest pattern !!!
  1. 36 * 1.75 + 24 * 3.25 = ?
    A. 146
    B. 141
    C. 143
    D. 142
    E. None of these
    Answer & Explanation
    Answer – B. 141
    Explanation :
    63 + 78 = 141
  2. 16.23 * 12.9 + 17.32 =  ?
    A. 224.87
    B. 234.87
    C. 244.87
    D. 226.87
    E. None of these
    Answer & Explanation
    Answer – D. 226.87
    Explanation :
    16.23 * 12.9 = 209.367
    209.367 + 17.32 = 226.87
  3. 444 ÷ (16*1.5) = ?
    A. 14.5
    B. 16.5
    C. 18.5
    D. 20.5
    E. None of these
    Answer & Explanation
    Answer – C. 18.5
    Explanation :
    16 * 1.5 = 24; 444 ÷ 24 = 18.5
  4. ? / 529 = 324 / ? 
    A. 325
    B. 414
    C. 456
    D. 448
    E. 352
    Answer & Explanation
    Answer – B. 414
    Explanation :
    x² = 529 * 324 => x = 23 * 18 = 414
  5. [18 * 15 – 20] / [(3.2 + 9.4) – 7.6] = ?
    A. 30
    B. 40
    C. 50
    D. 70
    E. 60
    Answer & Explanation
    Answer – C. 50
    Explanation :
    250 / 5 = 50
Decimal fraction

Decimal fraction

Welcome to aptitude tricks
Main Concepts and Results
• A fraction is a number representing a part of a whole. This whole
may be a single object or a group of objects.
• A fraction whose numerator is less than the denominator is called a
proper fraction, otherwise it is called an improper fraction.
• Numbers of the type
5 4 1 3 ,8 ,2
7 9 5 etc. are called mixed fractions
(numbers).
• An improper fraction can be converted into a mixed fraction and
vice versa.
• Fractions equivalent to a given fraction can be obtained by
multiplying or dividing its numerator and denominator by a non-
zero number.
• A fraction in which there is no common factor, except 1, in its
numerator and denominator is called a fraction in the simplest or
lowest form.
• Fractions with same denominators are called like fractions and if
the denominators are different, then they are called unlike fractions.
• Fractions can be compared by converting them into like fractions
and then arranging them in ascending or descending order.
• Addition (or subtraction) of like fractions can be done by adding
(or subtracting) their numerators.

Boats and stream 3

Boats and stream 3

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1) A man can row 18km/hr in still water. speed of the man in downstream is thrice the speed in upstream. Find the rate of stream.
Solution:
Let, Speed of man in upstream a = a
Speed of man in downstream b = 3a
Speed of man in still water u = ½(a + b)
Speed of man in still water = ½(3a + a) = 2a
We know speed of man in still water = 18
So, a = 9
Rate of stream = ½(27 – 9) =9 km/hr
2) A boat can cover certain distance in downstream in 1hr. and it takes 1½hr to cover same distance in upstream. If speed of the stream is 3kmph, then what will be the speed of boat?
Solution:
let speed of boat in still water be x
speed in downstream = x + 3
speed in upstream = x – 3
Since boat covered same distance in upstream and downstream,
(x + 3)*1 = (x – 3)*(3/2)
speed in downstream x= 15 kmph
3) A man can row three quarter of a km against the stream in 11¼ min and down the stream in 7½min. what is the speed of man in still water.
Solution:
Speed of man in Upstream = ((¾)/(45/4)) *60
= 4 kmph
Speed of man in Downstream = ((¾)/(15/4))* 60
= 12 kmph
Speed of man in still water u = ½(a + b)
Speed of man in still water = (12 + 4)*½ = 8 kmph
4) A streamer takes 3hr to cover a distance of 24km upstream, if the rate of stream is 3 kmph. Then find the speed of streamer in still water.
Solution:
Upstream speed = 24/3 =8kmph
Rate of stream = 3kmph
Speed of streamer – rate of stream = upstream
Speed of streamer = 11kmph
5) The distance between two points is 36km. A boat rows in still water at 6kmph, it takes 8hr less to cover dist in downstream in comparison to that in upstream. Find the rate of stream.
Solution:
Time = distance * speed
Difference between time taken to cover upstream and downstream is 8hr
(36/(6-x)) – (36/(6+x)) = 8
36(6+x) – 36(6-x) = 8(36-x2)
9x = 36- x2
x2+9x-36 = 0 (to find value of x use quadratic equation technique)
(x+12)(x-3)=0
Speed cannot be negative so x = 3kmph
Rate of stream = 3kmph
Boats and stream 2

Boats and stream 2

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Basic Concepts of Questions on Boats and Streams

  1. A boat is said to go downstream, if the boat goes in the direction of stream.
  2. A boat is said to go upstream, if the boat goes opposite to the direction of stream.

Basic Formulas

  1. If speed of boat in still water is b km/hr and speed of stream is s km/hr,
    • Speed of boat in downstream = (b  + s) km/hr , since the boat goes with the stream of water.
    • Speed of boat in upstream = (b  - s) km/hr. The boat goes against the stream of water and hence its speed gets reduced.

Shortcuts With Explanation

Scenario 1: Given a boat travels downstream with speed d km/hr and it travels with speed ukm/hr upstream. Find the speed of stream and speed of boat in still water.
Let speed of boat in still water be bkm/hr and speed of stream be skm/hr.
Then b + s  = d and b – s = u.
Solving the 2 equations we get,
b = (d + u)/2
s = (d – u)/2

Scenario 2: A man can row a boat, certain distance downstream in td hours and returns the same distance upstream in tu hours. If the speed of stream is s km/h, then the speed of boat in still water is given by
We know distance = speed * time
Let the speed of boat be b km/hr
Case downstream:
    d = (b + s) * td
Case upstream:
    d = (b - s) * tu

=>    (b + s) / (b - s) = tu / td

b = [(tu + td) / (tu - td)] * s

Scenario 3: A man can row in still water at bkm/h. In a stream flowing at  s km/h, if it takes him t hours to row to a place and come back, then the distance between two places, d is given by
Downstream:  Let the time taken to go downstream be td
    d = (b + s) * td

Upstream: Let the time taken to go upstream be tu
    d = (b - s) * tu

td + tu = t
[d / (b + s)] + [d / (b - s)] = t
So, d = t * [(b2 - s2) / 2b]
OR
d = [t * (Speed to go downstream) * (Speed to go upstream)]/[2 * Speed of boat or man in still water]

Scenario 4: A man can row in still water at bkm/h. In a stream flowing at s km/h, if it takes t hours more in upstream than to go downstream for the same distance, then the distance d is given by
Time taken to go upstream = t + Time taken to go downstream
(d / (b - s)) = t + (d / (b + s))
=> d [ 2s / (b2 - s2 ] = t
So, d = t * [(b2 - s2) / 2s]
OR
d = [t * (Speed to go downstream) * (Speed to go upstream)] / [2 * Speed of still water]
Boats and stream 1

Boats and stream 1

Welcome to aptitude tricks
Aptitude questions on boats and streams are common in most of the aptitude tests for companies like Infosys, TCS, Wipro, HCL etc. The advantage with questions on boats and streams are that there are only two basic concepts behind them, and you can solve any questions with these concepts. This article explains how to solve boats and streams problems easily and quickly. The types of questions that can be expected from quantitative aptitude section of boats and streams include following
  • You will be given the speed of boat in still water and the speed of stream. You have to find the time taken by boat to go upstream and downstream.
  • You will be given the speed of boat to go up & down the stream, you will be asked to find speed of boat in still water and speed of stream
  • You will be given speed of boat in up and down stream and will be asked to find the average speed of boat.
  • You will be given the time taken by boat to reach a place in up and downstream and will be asked to find the distance to the place
The key point here is you can solve any of these questions using the formulas and short cuts given below.
Profit and loss 2

Profit and loss 2

Welcome to aptitude tricks
Question 6: A dishonest merchant sells his grocery using weights 15% less than the true weights and makes a profit of 20%. Find his total gain percentage.
Solution:
Let us consider 1 kg of grocery bag. Its actual weight is 85% of 1000 gm = 850 gm.
Let the cost price of each gram be Re. 1. Then the CP of each bag = Rs. 850.
SP of 1 kg of bag = 120% of the true CP
Therefore, SP = 120/100 * 1000 = Rs. 1200
Gain = 1200 – 850 = 350
Hence Gain % = 350/850 * 100 = 41.17%

Question 7: A man bought two bicycles for Rs. 2500 each. If he sells one at a profit of 5%, then how much should he sell the other so that he makes a profit of 20% on the whole?
Solution:
Before we start, it’s important to note here that it is not 15% to be added to 5% to make it a total of 20%.
Let the other profit percent be x.
Then, our equation looks like this.
105/100 * 2500 + [(100+x)/100] * 2500 = 120/100 * 5000 → x= 35.
Hence, if he makes a profit of 35% on the second, it comes to a total of 20% profit on the whole.

Question 8: A shopkeeper allows a discount of 10% on the marked price and still gains 17% on the whole. Find at what percent above the cost price did he mark his goods.
Solution:
Let the cost price be 100. Then SP = 117.
Let the marked price be x.
So, 90% of x = 117 → x = 130.
Therefore, he marked his goods 30% above the cost price.

Question 9: A shopkeeper offers a discount of 20% on the selling price. On a special sale day, he offers an extra 25% off coupon after the first discount. If the article was sold for Rs. 3600, find
  1. The marked price of the article and
  2. The cost price if the shopkeeper still makes a profit of 80% on the whole after all discounts are applied.
Solution:
Let the marked price of the article be x.
First a 20% discount was offered, on which another 25% discount was offered.
So, 75% of 80% of x = 3600
75/100 * 80/100 * x = 3600 → x = 6000.
So the article was marked at Rs. 6000.
Cost price of the article = [100/(100+80)]*3600 = Rs. 2000.
It is important to note here that this DOES NOT equal to a 45% discount on the whole. When different discounts are applied successively, they CANNOT be added.
Number system

Number system

Every placement test on quantitative aptitude will contain at least 30% questions on number systems and number series. Aptitude questions on number system form the backbone for placement preparation. You can score easily on quantitative aptitude section if you understand the basics of number system. Since the questions on number systems are simple, importance lies in acquiring the right skills to tackle these problems with speed.
Practicing problems on numbers systems not only helps in improving your speed but also provides a strong base for solving other quantitative aptitude sections like HCF and LCMaveragespercentagestime and speedpipes and cisterns etc as well. In this tutorial let's look at how to solve number system problems easily nd quickly.




Numbers – Aptitude Test Questions, Tricks & Shortcuts

    131 Votes

Every placement test on quantitative aptitude will contain at least 30% questions on number systems and number series. Aptitude questions on number system form the backbone for placement preparation. You can score easily on quantitative aptitude section if you understand the basics of number system. Since the questions on number systems are simple, importance lies in acquiring the right skills to tackle these problems with speed.
Practicing problems on numbers systems not only helps in improving your speed but also provides a strong base for solving other quantitative aptitude sections like HCF and LCMaveragespercentagestime and speedpipes and cisterns etc as well. In this tutorial let's look at how to solve number system problems easily and quickly.
Numbers are fun to learn. If you learn the concepts thoroughly you will find that solving aptitude questions on number system is a cake walk for you. There are lot of concepts involved and hence even a simple question might look a bit too complex or trickier to solve.
We at a4academics will provide you with right tools to tackle quantitative tests on number system. Below is the list of important formulas on number system and tips to help you understand and prepare for the quantitative aptitude questions on number systems.
Before learning the tips and tricks, please go through the tutorial given below to understand the number system math concepts in detail. This tutorial will enhance your problem solving skills.
Number System Tutorial
Part I: Integers, Fractions, Prime and Composite
Part II: Divisibility, Remainder, HCF and LCM
Part III: Factors, Multiples, Unit Digit and Last Two Digits of Exponents

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.