168E- RATIO AND PROPORTIONS

Ratio and Proportion Methods shortcut tricks

Shortcut Tricks are very important things in competitive exam. Competitive exams are all about time. If you manage your time then you can do well in those exams. Most of us miss that part. We provide examples on Ratio and Proportion shortcut tricks here in this page below. These shortcut tricks cover all sorts of tricks on Ratio and Proportion. Visitors are requested to carefully read all shortcut examples. These examples here will help you to better understand shortcut tricks on ratio and proportion.
First of all do a practice set on math of any exam. Write down twenty math problems related to this topic on a page. Do first ten maths using basic formula of this math topic. You also need to keep track of the time. After solving all ten math questions write down total time taken by you to solve those questions. Now go through our page for ratio and proportion shortcut trick. After this do remaining ten questions and apply shortcut formula on those math problems. Again keep track of the time. This time you will surely see improvement in your timing. But this is not enough. You need more practice to improve your timing more.
You all know that math portion is very much important in competitive exams. That doesn’t mean that other sections are not so important. But only math portion can leads you to a good score. A good score comes with practice and practice. All you need to do is to do math problems correctly within time, and only shortcut tricks can give you that success. But it doesn’t mean that without using shortcut tricks you can’t do any math problems. You may have that potential that you may do maths within time without using any shortcut tricks. But so many people can’t do this. Here we prepared ratio and proportion shortcut tricks for those people. Here in this page we try to put all types of shortcut tricks on Ratio and Proportion. But we may miss few of them. If you know anything else rather than this please do share with us. Your little help will help so many needy.
Now we will discuss some basic ideas of Ratio and Proportion. On the basis of these ideas we will learn trick and tips of shortcut ratio and proportion. If you think that how to solve ratio and proportion questions using ratio and proportion shortcut tricks, then further studies will help you to do so.

What is Ratio?: A ratio is a relationship between two numbers by division of the same kind. The ration of a to b is written as a : b = a / b, In ratio a : b, we can say that a as the first term or antecedent and b the second term or consequent.

Example :  The ratio 4 : 9 we can represent as  4 / 9 after this 4 is a antecedent and, consequent = 9

Rule of ratio :  In ratio multiplication or division of each and every term of a ratio by the same non- zero number does not affect the ratio.

Different type of ratio problem are given in Quantitative Aptitude which is a very essential topic in banking exam. Under below given some more example for your better practice.
Anything we learn in our school days was basics and that is well enough for passing our school exams. Now the time has come to learn for our competitive exams. For this we need our basics but also we have to learn something new. That’s where shortcut tricks and formula are comes into action.

What is Proportion?
The idea of proportions is that two ratios are like equal.
If a : b = c : d, we write a : b : : c : d,
Ex. 3 / 15 = 1 / 5
a and d called extremes, where as b and c called mean terms.

Proportion of quantities
the four quantities like a, b, c, d we can say proportion then we can express it
a : b = c : d
Then a : b : : c : d  <–> ( a x d ) = ( b x c )
product of means = product of extremes.

If there is given three quantities like a, d, c of same like then we can say it proportion of continued.
a : d = d : c , d is called mean term. a and c are called extremes.

Example of Ratio and Proportion method :


Example Set -1

Ratio Example 1:

Ratio Example 1 shortcut tricks and formula based problem are very important for Competitive exams here is some problems which are given in exams that is some item or product are divide into persons and find the number of item that the person should have, we discuss this example in ratio example 1.
This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam. Under below  given some more example for your better practice.

Example 1: If A : B = 7 : 9, C : D = 6 : 13 then A : B : C is
Answer : A : B : C = (7 x 6) : (9 x 6) : (9 x 13) = 42 : 54 : 117

Example 2:
If P : Q : R = 2 : 3 : 4, Then P / Q : Q / R : R / P = ?
Answer :
P : Q : R = 2 : 3 : 4.
Let P = 2k,
Q = 3k,
R = 4k.
Then,
P / Q = 2k / 3k = 2 / 3 ,
Q / R = 3k / 4k = 3 / 4
R / P = 4k / 2k = 2 / 1.
At first we do LCM of 2, 3, 4 is 12, now we multiplied with ratio numbers ,
Like, 2 x 12 / 3 = 8
3 x 12 / 4 = 9
2 x 12 / 1 = 24
So,the ratio of P / Q : Q / R : R / P is 8 : 9 : 24.

Example 3:
If a : b = 3 : 7 and b : c = 5 : 9, Find a : b : c.
Answer: 
A : B = 3 : 7,
B : C = 5 : 9
= ( 5 x 7 / 5 ) : ( 9 x 7 / 5 ) = 7 : 63 / 5.
= A : B : C = 3 : 7 : 63 / 5 = 15 : 35 : 63 .

Example 4:
If P : Q = 2 : 3 and Q : R = 4 : 5, then R : P = ?
Answer :
P / R = ( P / Q x Q / R ) = ( 2 / 3 x 4 / 5 ) = 8 / 15
=R / P = 15 / 8 = R : P = 15 : 8.

Example 5:
If P : Q = 2 : 3, Q : R = 4 : 5 and R : S = 6 : 7, Then P : S = ?
Answer :
P / S = ( P / Q x Q / R x R / S ) = ( 2 /3 x 4 / 5 x 6 / 7 ) = 16 / 35 = P : S = 16 : 35.

Example 6:
If P : Q = 4 : 3 , Q : R = 3 : 5 and R : S = 10 : 9, Then P : Q : R : S = ?
Answer :
P : Q = 4 : 3
Q : R = 3 : 5
R : S =10 : 9
= 4 x 3 x 10 : 3 x 3 x 10 : 3 x 5 x 10 : 3 x 5 x 9
P : Q : R : S = 120 : 90 : 150 : 135 = 8 : 6 : 10 : 9.

Example 7: what would be the 3rd proportional to 0.25 to 0.38 ?
Answer : 0.25 : 0.38 :: 0.38 : X
X = 0.38 x 0.38 / 0.25 = 0.5776

Eaxmple 8: Divide the Rs, 520 in the ratio of 6 : 4 in between ramesh and suresh, How much amount would be both are getting ?
Answer: Sum of ratio = ( 6 + 4 )= 10,
Ramesh got his amount = 520 x 6 / 10 = 312.
Suresh got his amount = 520 x 4 / 10 = 208.

Example 9:
What is the smallest part, If 75 is divided into three parts proportional to 3, 5, 8, 9.
Answer :
ratio is = 3 : 5 : 8 : 9, sum of ratio terms is = 25.
So the smallest part is = ( 75 x 3 / 25 ) = 9.

Example 10:
Rama gives his pencils between his four friends Rakesh, Rahul, Ranjan, and Rohit in the ratio 1 / 2 : 1 / 3 : 1 / 4 : 1 / 5. What would be the minimum number of pencils Rama should have?
Answer :
Rakesh : Rahul : Ranjan : Rohit = 1 / 2 : 1 / 3 : 1 / 4 : 1 / 5
Step 1: At First we need to do is LCM of 2,3,4 and 5 is 60.
Step 2: Then pencils are given in ratio among friends,
Rakesh = ( 1 / 2 x 60 ) = 30.
Rahul = ( 1 / 3 x 60 ) = 20.
Ranjan = ( 1 / 4 x 60 ) = 15.
Rohit = ( 1 / 5 x 60 ) = 12.
Step 3: Total number of pencils are ( 30 x + 20 x + 15 x + 12 x) = 77 x.
The minimum number of pencils are  x = 1.
Rama should have only 77 pencils.

Example 11:
Two numbers are respectively 40% and 60% more than third number. What would be the ratio of two numbers ?
Answer :
Step 1: Let the third number is A
Then first number is 140% of A = 140 x A / 100 = 7A / 5 and second number is 160% of B = 160 x B / 100 = 8B / 5.
Step 2: now ratio of first and second number is 7A / 5 : 8B / 5 = 35A : 40B = 7 : 8.

Example 12: A sum money is divided among P, Q, R, S in the ratio of 2 : 3 : 1 : 7 respectively. If the share of Q is Rs.9872, than what would be the total amount of P and S together ?
Answer: Share of Q = 9872, P, Q, R, S in the ratio of 2 : 3 : 1 : 7
Total amount of P ans S together is = 9872 x 9 = 88848.

Example 13:
If A : B = 4 : 9 and B : C = 3 : 6 , Then find A : C is :
Answer : (A / B = 4 / 9, B / C = 3 / 6 )
= A / C = ( A /B x B / C) = ( 4 / 9 x 3 / 6 ) = 2 / 9 = A : C = 2 : 9.

Example Set -2


Example 1: In a town, 20% of the mans are same in numbers as 1 / 4th of the women. What would be the ratio of mans and woman in that town ?
Answer : 20% mans
= 1 / 4 womens<=> 20man / 100
So, mans = 5 / 4 womens
mans / womens = 5 / 4
mans : womens = 5 : 4

Example 2: In a school, the number of ratio of boys and girls is 4 : 9, after inclusion of 32 new girls, the ratio become 4 : 17. How many boys were present at starting in this school ?
Answer : 4x / 9x + 32 = 4 / 17
68x = 36x + 128
x = 4
So, the number of boys in the school is (4 x 4) = 16.

Example 3: Ina school ratio of number of boys and girls is 9 : 6 and if there present 180 boys, than find the total number of students in the school.
Answer: Let number of boys and girls 9x and 6x.
Then 9x = 180 = x = 20.
So, Total number of students is = 15x = (15 x 20) = 300.

Example 4:
Share Rs.4200 among joy, sanjay and bijoy in the ration 2 : 4 : 6.Find the amount received by sanjay.
Answer :
Amount received by sanjay.
4 / 12 X 4200 = 1400= ( related ratio / sum of ratio ) x Total amount
So, the Amount received by sanjay is 1400.

Example 5:
Find the mean proportional between given two number that is 64 and 49.
Answer :
The mean proportion of two numbers is
Root of 64 and 49 is √8 x √ 7 = 8 x 7 = 56.
So, the mean proportional is 56.

Example 6:
Rs. 385 were divided among P , Q , R in such a way that P had Rs 20 more than Q and R had Rs 15 more than P . How much was R’s share ?
Answer :
Let Q gets Rs x. Then We can say P gets Rs (x + 20 ) and R gets Rs ( x + 35) .
x + 20 + x + x + 35 = 385
3x = 330
x = 110 .
R’s share = Rs ( 110 + 35 ) = Rs 145.

Example 7:
Rs 1210 were divided among three person P, Q, R so that P : Q = 5 : 4 and Q : R = 9 : 10. Then R gets the amount.
Answer :
P : Q = 5 : 4, Q : R = 9 : 10 = ( 9 x 4 / 9 ) : ( 10 x 4 / 9 ) = 4 : 40 / 9.
So, P : Q : R = 5 : 4 : 40 /9 = 45 : 36 : 40
Sum of ratio terms is = ( 45 + 36 + 40 ) =121.
R share of amount is Rs (1210 x 40 / 121) = Rs. 400 .

Example 8:
Rs 64000 are divided among three friends in the ratio 3 / 5 : 2 / 1 : 5 / 3 . The share of the second friend is :
Answer :
Here the given ratio is = 3 / 5 : 2 / 1 : 5 / 3 = 9 : 30 : 25 .
So the Share of the second friend = Rs ( 64000 x 30 / 64) = 30000.

Example 9:
Which of the following ratio is the greatest ?
7 : 15 , 15 : 23 , 17 : 25 , 21 : 29
Answer :
7 / 15 = 0.466
15 / 23 = 0.652
17 / 25 = 0.68
21 / 29 = 0.724
So, 0.724 is greatest and therefore, 21 : 29.

Example 10:
What number has to be added to each term of 3 : 5 to make the ratio 5 : 6 :
Answer :
Let the number to be added x , Then
3 + x / 5 + x = 5 / 6
6 ( 3 + x ) = 5 ( 5 + x )
x = ( 25 – 18 ) = 7
So , the number to be added is 7.

Example 11:
On dividing a sum of Rs 832 between Paul and john in the ratio 3 : 5 , their shares are
Answer :
So the given ratio is = 3 : 5
So , the 1st part = Rs ( 832 x 5 / 8 ) = Rs 520 , 2nd part = Rs ( 832 x 3 / 8 ) = Rs 312
So , their shares are :  312 and 520.

Example 12:
A certain amount was divided between A and B in the ratio 4 : 3 . If B’s share was Rs, 4800, the total amount was ?
Answer :
If B ‘s share is Rs, 3, total amount = Rs, 7
If B’s share is Rs, 4800, total amount  = Rs, (7  / 3 x 4800) = 11200.

Example 13:
The Salary of Three friend A, B, C are divided into ratio 5 : 6 : 8. If the increment has given of 10%, 20%, 25%, Find the new ratio of three friend salary ?
Answer :
Step 1: We assume ration as 5x, 6x, 8x
now the increment of new salary of  A is 10% = 110 / 100, B is 20% = 120 / 100, C is 25% = 125 / 100.
Step 2: A,s new salary is 110 x 5X / 100 = 55X / 10.
B,s new salary is 120 x 6X / 100 = 36X / 5.
C,s new salary is 125 x 8X / 100 = 10.
Step 3: New ratio is 55X / 10 : 36X / 5 : 40X / 4.


Example Set -3

Example 1:
A money bag contains 50 p, 25 p, and 10 p coins in the ratio 5 : 9 : 4,  and the total amounting to Rs.206.
Find the individual number of coins of each type.
Answer :
Step 1: Let the number of 50 p ,25 p, and 10 p coins be 5x, 9x, 4x respectively.
we can say two 50 paise = 1 rupee that’s why we take 2
we can say four 25 paise = 1 rupee that’s why we take 4
we can say Ten 10 paise = 1 rupee that’s why we take 10
Then, 5x / 2 + 9x / 4 + 4x / 10 = 206
= 50x + 45x + 8x = 4120
= 103x = 4120
= x = 40.

Step 2: Number of 50 p coins is ( 5 x 40 = 200 ),
Number of 25 p coins is( 9 x 40 = 360 ),
Number of 10 p coins ( 4 x 40 = 160 ),
Example 2:
On a self there are 4 books on Economics, 3 books on Management and 4 books on Statistics. In how many different ways can be the books be arranged so that the books on Economics  are kept together ?
Answer :
( 4 books on Statistics ! + 3 books on Management ! + 1 x   4 books on Economics ! )
Total ways = 8! x 4!
= ( 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 ) x ( 4 x 3 x 2 x 1 )
= 40320 x 24
=967680.
So , we can 967680 way be the books be arranged.

Example 3:
How many bags are required for filling 1824 kg of wheat if each bag filled with 152 kg of wheat ?
Answer :
Each bag filled means 1 bag filled with 152 kg  .
Total kg of wheat is 1824 kg , So we divide it  by each bag filled with 152 kg
Number of bags = 1824 / 152
= 12.

Example 4:
An urn contains 4 green , 5 blue , 2 red  and 3 yellow marbles. If four marbles are drawn at random, what is the probability that two are blue and two are red ?
Answer :
Required probability = ⁵C₂ x ²C₂ / ¹⁴C₄
= 10 / 1001.

Example 5:
An urn contains 4 green , 5 blue , 2 red  and 3 yellow marbles. If eight marbles are drawn at random, what is the probability that there are equal number of marbles of each color ?
Answer :
Required probability = ⁴C₂ x ²C₂ x ³C₂ / ¹⁴C₈
= 18 / 3003.

Example 6:
An urn contains 4 green , 5 blue , 2 red  and 3 yellow marbles. If two marbles are drawn at random, what is the probability that both are red or at least one is red ?
Answer :
Required probability = ²C₂ / ¹⁴C₂ + [ 1 – ¹²C₂ / ¹⁴C₂]
= 1 / 91 + [ – 66 / 91 ]
= 1 / 91 + 25 / 91
= 26 / 91.

Example 7:
An urn contains 4 green , 5 blue , 2 red  and 3 yellow marbles. If three marbles are drawn at random, what is the probability that at least one is yellow ?
Answer :
Required probability = 1 – ¹¹C₃ / ¹⁴C₃
= 1 – 165 / 364
= 199 / 364.

Example 8:
An urn contains 4 green , 5 blue , 2 red  and 3 yellow marbles. If three marbles are drawn at random, what is the probability that none is green ?
Answer :
Required probability = ¹⁰C₃ / ¹⁴C₃
= 30 / 91.

Example 9: In a cotton Bag has 50 paise and 25 paise coins, and the cotton bag contains total 60 coins which is sum of Rs.24.25. In cotton bag How many 25 paise coins are present ?
Answer: X x 0.50 + ( 60 – X ) x 0.25 = 24.25
0.50X + 15 – 0.25X = 24.25
0.25X = 24.25 – 15
0.25X = 9.25
X = 925 x 100 / 25 x 100
X = 37
Total no of coins is 60. So, ( 60 – X ) = ( 60 – 37 ) = 23
So, 25 paise coins are present in bag is 23.

Example 10:
If each one plastic bag is filled with 160 Kgs of rice, than for 2240 Kgs of rice how many plastic bags are required ?
Answer :Total rice is 2240, So required bags is 2240 / 160 = 14.
So plastic bags are required 14.


Example Set -4
Ratio based problem are very important for Competitive exams. Here is some problems we provide you which are given in exams that is, some item are divide into persons and find the amount of one or two persons, we discuss this example in ration example 4. This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam. Under below given some more example and ratio shortcut tricks for your better practice.
Example 1: In the ratio of 8 : 12 Copper and silver are melted together. What is the weight of melted mixture if 24 kg of silver has been consumed in it ?
Answer: The mixture melted of copper and silver = (12 + 8)kg = 20 kg.
for 24kg silver, so mixture melted is = 20 x 24 / 8 = 60 kg.
 
Example 2:
Rs 75,500/- are divided between A and B in the ratio 1 : 3. what is the difference between thrice the share of A and twice the share of B ?
Answer :
Share of A = 75,500 x 1  / 1 + 3 = 75,500 x 1 / 4 = 18875 .
Share of B = 75,500 x 3 / 1 + 3 = 75,500 x 3 /4  = 56625 .
Difference between thrice the share of A and twice the share of B is
= 2B – 3A
= 2 x 56625 – 3 x 18875
= 113250 – 56625
= 56625.

Example 3:
Milk and water in the ratio 5 : 3 is contain in a 20 litres of mixture. If 4 litres of this mixture be replaced by 4 litres of milk, the ratio of milk to water in the new mixture would be ?
Answer :
Quantity of milk in the mixture after taking out 4 liters of the mixture = ( 16 x 5 / 8 ) = 10 litres.
Now we add 4 liters of milk in this mixture.
quantity of milk in 20 litres of new mix = ( 10 + 4 ) = 14 litres.
quantity of water in it ( 20 – 14 ) = 6 litres.
Ratio of milk and water in the new ratio mix is = 14 : 6 = 7 : 3.

Example 4:
Rs 75,500/- are divided between A and B in the ratio 1 : 3. what is the difference between thrice the share of A and twice the share of B ?
Answer :
Share of A = 75,500 x 1  / 1 + 3 = 75,500 x 1 / 4 = 18875 .
Share of B = 75,500 x 3 / 1 + 3 = 75,500 x 3 /4  = 56625 .
Difference between thrice the share of A and twice the share of B is
= 2B – 3A
= 2 x 56625 – 3 x 18875
= 113250 – 56625
= 56625.

Example 5:
In a bottle mixture of 80 liters and the ratio of milk and water is 3 : 2. If this mixture ratio is to be 2 : 3. What the quantity of water to be further added ?
Answer :
Step 1: Quantity of Milk ( 80 x 3 / 5 ) = 48 liters, So Quantity of water in it  ( 80 – 48 ) = 32 liters.
Step 2: New Ratio required 2 : 3, Let x water to be added, Then Milk : Water is = 48 : (32+x)
=48 / (32 + x).
Step 3: Now 48 / (32 + x) = 2 : 3
48 / (32 + x) = 2 / 3
2x = 144 – 64
x = 80/2
=40 liters.

Example 6:
The ratio between the number of men and women in a society is 31 : 23, When 75 more women are added in the society, this ratio becomes 124 : 107. How many more women should be added in the society in order to make the number of men and women be equal?
Answer :
31x / 23x + 75 = 124 / 107
3317x = 2852x + 9300
465x = 9300
x = 20 .
So , the number of women in society after added
= 20 x 23 + 75 = 535 .
So , the number of men in society after added
= 31 x 20 = 620 .
Number of more women is = 620 – 535 = 85.

Example 7: In a liquid mixture of 60 litres , the ratio of milk and water is 2 :1. If this ratio is to be 1 : 2, then the quantity of water to be further added is :
Answer :
So Quantity of milk = ( 60 x 2 / 3 ) = 40 litres.
Water in it = ( 60 – 40 ) = 20 litres .
new ratio required = 1 : 2 .
Let quantity of water to be added further be x liters
milk : water = 40 / ( 20 + x )
Now,
40 / ( 20 + x ) = 1 / 2
20 + x = 80
x = 60 lires
So, 60 litres of water to be added further.

Example 8:
20% alcohol present in a 15 litres mixture and rest of water. If 3 litres of water be mixed with it, the percentage of alcohol in the new mixture would be  ?
Answer :
20% present in a 15 litres mixture
So, 20 x 15 / 100 = 3 litres.
Water in it = ( 15 – 3 ) 12 litres.
New quantity of mixture = ( 15 + 3 ) = 18 litres.
Percentage of alcohol in new mix is = ( 3 x 100 / 18 ) % = 50 / 3.

Example 9:
A liquid mixture contain alcohol and water in the ratio of 4 : 3. If 5 liters of water is added to the mixture the ration becomes 4 : 5. Find the quantities of alcohol in the given mixture ?
Answer :
Let the quantity of alcohol and water be 4x and 3x liters.
Then,
4x / 3x + 5 = 4 / 5
= 20x = 4(3x + 5)
= 8x = 20
= x = 2.5
Quantity of alcohol = ( 4 x 2.5 ) = 10 liters.

Example 10: If the ratio of milk and water is 3 : 2 in a mixture of 80 liters. If this ratio later to be 2 : 3, Then what quantity of water to be added more.
Answer : Total mixture is 80 liters and ratio is 3 : 2
So, milk is 80 x 3 / 5 = 240 / 5 = 48
Water is ( 80 – 48 ) = 32
48 / ( 32 + x ) = 32
2x = 80
x = 40
Quantity of water to be added more is 40 liters.


Example 11: In a mixture of 25 litres has contains 40% milk and remaining is water, and if 5 litrs of water mix with it, then what would be the new percentage of milk in mixture ?
Answer :  40% milk means 60% water
So, milk in mixture is = 25 x 40 / 100 = 10 litres.
and water in mixure is ( 25 – 10 ) = 15 litres.
new mixture quantity = ( 25 + 5 ) = 30 litres.
10 x 100 / 30 = 100 / 3 litres.
new percentage of milk in mixture is 100 / 3 litres.

Now, you can solve the following Questions out of Test Paper No168-E23 and submit answers in the form of 1a,2b,3c,4d,5e,6b in the Comments Section. You will get your scorecard, answer key, and detailed explanation of each question. Please submit your email address, WhatsApp# to download full Practice Test Paper and similar other practice test papers. You can also download Lessons and Test Papers from https://www.facebook.com/groups/NextGenCareers/

QUANTITATIVE APTITUDE   ::   RATIO AND PROPORTIONS

TEST No:NGC-168-QA-E023 :: No of Questions:023   ::   Time Allowed : 18  Minutes
Directions: Please Tick Appropiate Option and Submit, Inform Serial Number of Questions where you need detailed Explanations to bhagirathprayash@gmail.com or Massage to 9462900411

Question 1:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

Pinku, Rinku and Tinku divide an amount of Rs 4,200 amongst themselves in the ratio of 7:8:6 respectively. If an amount of Rs 200 is added to each of their shares, what will be the new respective ratio of their shares of amount?
(a) 8 : 9 : 6 
(b) 7 : 9 : 5 
(c) 7 : 8 : 6 
(d) 8 : 9 : 7  

Question 2:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

The total number of students in a school is 3250. If the number of girls in the school is 1495, then what is the respective ratio of the total number of boys to the total number of girls in the school?
(a) 23 : 27 
(b) 25 : 29 
(c) 27 : 23 
(d) 29 : 25  

Question 3:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

Rinku and Pooja started a business initially with Rs 5,100 and Rs 6,600 respectively. If the total profit is Rs 2,730 what is Rinku’s share in the profit?
(a) Rs 1,530 
(b) Rs 1,540 
(c) Rs 1,200 
(d) Rs 1,190  

Question 4:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

A sum of money is to be divided equally amongst A, B, and C in the respective ratio of 3:4:5 and another sum of money is to be divided between E and F equally. If F got Rs 1050 less than A, how much amount did B receive?
(a) Rs 750 
(b) Rs 2000 
(c) Rs 1500 
(d) निर्धारित नहीं किया जा सकता Cannot be determined  

Question 5:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

The total number of boys in a school is 15% more than the total number of girls in the school. What is the respective ratio of the total number of boys to the total number of girls in the school?
(a) 17 : 23 
(b) 24 : 11 
(c) 23 : 20 
(d) 11 : 24  

Question 6:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

Which number should replace both the question marks in the following equation?
?/84 =189/?
(a) 126 
(b) 124 
(c) 130 
(d) 132  

Question 7:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

If A : B : C = 2 : 3 : 4,
then A/B : B/C : C/A is equal to :
(a) 4 : 9 : 16 
(b) 8 : 9 : 12 
(c) 8 : 9 : 16 
(d) 8 : 9 : 24  

Question 8:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs 1000 more than D, what is B’s share?
(a) Rs 500 
(b) Rs 1500 
(c) Rs 2000 
(d) Rs 2500  

Question 9:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

Seats for Maths, Physics and Biology are in the ratio of 5 : 7 : 8 respectively. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the respective ratio of increased seats?
(a) 2 : 3 : 4 
(b) 6 : 7 : 8 
(c) 6 : 8 : 9 
(d) Cannot be determined  

Question 10:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

Ratio of the earning of A and B is 4:7 respectively. If the earnings of A increases by 50% and the earnings of B decreases by 25% the new ratio of their earnings becomes 8:7 respectively. What are A’s earnings?
(a) Rs 26,000 
(b) Rs 28,000 
(c) Rs 21,000 
(d) Data inadequate  

Question 11:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

In a college the students in Arts and Commerce faculties were in the ratio of 4 : 5 respectively. When 65 more students joined Commerce faculty the ratio became 8 : 11 respectively. How many students are there in Arts faculty?
(a) 520 
(b) 650 
(c) 715 
(d) Cannot be determined  

Question 12:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

The ratio between the boys and girls in a class is 6 : 5 respectively. If 8 more boys join the class and two girls leave the class then the respective ratio becomes 11 : 7. What is the number of boys in the class now?
(a) 28 
(b) 38 
(c) 44 
(d) 36  

Question 13:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

If ½ of Sunit’s salary is equal to ⅖ of Rajan’s salary and their total salary is Rs 36,000, find Rajan’s salary.
(a) Rs 16,000 
(b) Rs 20,000 
(c) Rs 22,000 
(d) Rs 14,00  

Question 14:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

The largest and the second largest angles of a triangle are in the ratio of 13 : 12 respectively. The smallest angle is 20% of the sum of the largest and the second largest angles. What is the sum of the smallest and the second largest angles?
(a) 120° 
(b) 108° 
(c) 100° 
(d) 102°  

Question 15:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

Mr. Pandit owned 950 gold coins all of which he distributed amongst his three daughters Lalita, Amita and Neeta. Lalita gave 25 gold coins to her husband, Amita donated 15 gold coins and Neeta made jewellery out of 30 gold coins. The new respective ratio of the coins left with them was 20 : 73 : 83. How many gold coins did Amita receive from Mr. Pandit?
(a) 380 
(b) 415 
(c) 400 
(d) 350  

Question 16:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

Salaries of A, B and C were in the ratio of 3 : 5 : 7 respectively. If their salaries were increased by 50%, 60% and 50% respectively, what will be the new ratio of their respective salaries?
(a) 3 : 6 : 7 
(b) 4 : 5 : 7 
(c) 9 : 16 : 21 
(d) Data inadequate  

Question 17:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

At Narmada Sarovar Bachao (NSB) demonstration, sup­porters of Ms Patkar outnumbered the police by 9 : 1. The police arrested 135 NSB supporters averaging 5 for every 3 policemen. How many supporters of NSB were there in the demonstration?
(a) 1215 
(b) 665 
(c) 405 
(d) 729  

Question 18:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

The scale of map is ¾ of cm = 1 km. If the distance on the map between two towns is 60 cm, then the actual distance is
(a) 60 km 
(b) 80 km 
(c) 75 km 
(d) 50 km  

Question 19:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

The ratio of the number of boys and girls in a school is 2 : 5. If there was 350 students in the school, find the number of girls in the school.
(a) 200 
(b) 225 
(c) 250 
(d) 275  

Question 20:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

The salaries of A,B,C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be the new ratio of their salaries?
(a) 3 : 3 : 10 
(b) 10 : 11 : 20 
(c) 23 : 33 : 60 
(d) Cannot be determined  

Question 21:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

In an express train, the passengers travelling in A.C. sleeper class, First class and Sleeper class are in the ratio 1:2:7, and rate for each class is in the ratio 5 : 4 : 2. If the total income from this train is Rs 54, 000, find the income of Indian Railways from A.C. sleeper class.
(a) Rs 12,000 
(b) Rs 20,000 
(c) Rs 22,000 
(d) Rs 10,000  

Question 22:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

What is the ratio whose terms differ by 40 and the measure of which is ²⁄₇?
(a) 16 : 56 
(b) 14 : 56 
(c) 15 : 56 
(d) 16 : 72  

Question 23:  NEXTGEN CAREER COACHING#9462900411--NGC-168-QA  

A photograph measuring 2¹⁄₂ × 1⁷⁄₈is to be enlarged so that the length will be 4 inch. How many inches will the enlarged breadth be?
(a) 1½ 
(b) 2⅛ 
(c) 3 
(d) 3⅜  

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163E - NUMBER SERIES

Number Series Methods shortcut tricks

Number Series shortcut tricks are very important thing to know for your exams. Time takes a huge part in competitive exams. If you manage your time then you can do well in those exams. Most of us miss that part. Few examples on number series shortcuts is given in this page below. These shortcut tricks cover all sorts of tricks on Number Series. We request all aspirants to read all examples carefully. These examples will help you to understand shortcut tricks on Number Series.
Now we will discuss some basic ideas of Number Series. On the basis of these ideas we will learn trick and tips of shortcut number series. If you think that how to solve number series questions using number series shortcut tricks, then further studies will help you to do so.
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What is Number Series ?
Number series is a form of numbers in a certain sequence, where some numbers are mistakenly put into the series of numbers and some number is missing in that series, we need to observe first and then find the accurate number to that series of numbers.
Anything we learn in our school days was basics and that is well enough for passing our school exams. Now the time has come to learn for our competitive exams. For this we need our basics but also we have to learn something new. That’s where shortcut tricks are comes into action.
In competitive exams number series are given and where you need to find missing numbers and mistakenly put into the series numbers. The number series are come in different types. At first you have to decided what type of series are given in papers then according with this you have to use shortcut tricks as fast as you can.
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Different types of Number Series
There are some format of series which are given in Exams.

Perfect Square Series
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Perfect square series is a arrangement of numbers in a certain order, where some numbers Series are based on square of a number which is in same order you need to place one square number that is missing in that given series, we need to observe and find the accurate number to the series of numbers.
This type of problem are given in Quantitative Aptitude which is a very essential in banking exam. It is simple to work on perfect square root numbers you can easily obtain the result of perfect square number. How you get easily by  Perfect square numbers  missing term by memorize square and square root numbers shortcut tricks .
The square of same  number and the square result of a number which is equal to the square of another same element. In mathematical world, a square number or perfect square is number of an integer positive integer that is the square of an same integer number always and the numbers are non-negative.
In other words, we say it is the result of  product of multiplication of some positive integer numbers with itself always. For example, we consider 4 is a result of square numbers, since it as 2 × 2 in normal way.
The normal representation of square numbers is n² and that is similar with products of n × n, but it is similar with exponentiation of n² ,
In Square numbers are positive number. So we can explain it that a positive number is a square number, where its square roots are always integers positive numbers. so For example, √4 = ±2, so 4 is a square number.
Here we see the some examples that how the perfect square are arranged how the missing square series are arranged.
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Example 1: 841, ?, 2401, 3481, 4761
Answer : 29², 39², 49², 59², 69²

Example 2: 1, 9, 25, ?, 81, 121
Answer : 1², 3², 5², 7², 9², 11²

Example 3: 289, 225, 169, ?, 81
Answer: 17², 15², 13², 11², 9²

Example 4: 441, 484, 529, 576, ?,
Answer: 441 = 21², 484 = 22², 529 = 23², 576 = 24² ,625 = 25².

Example 5: 121, 144, 169, ?, 225
Answer: 121 = 11², 144 = 12², 169 = 13², 196 = 14², 225 = 15².

Example 6: ?, 2116, 2209, 2304, 2401, 2500
Answer: 2025 = 45², 2116 = 46², 2304 = 48², 2401 = 49², 2500 = 50²

Example 7: 961, 1024, ?, 1156, 1225
Answer: 961 = 31², 1024= 32², 1089 = 33², 1156 = 34², 1225 = 35².

Example 8: 36, ?, 64, 81, 100, 121
Answer: 36 = 6², 49 = 7², 64 = 8², 81 = 9², 100 = 10², 121 = 11².

Example 9: 121 , 169 , ? , 289 , 361
Answer : 11² = 121 , 13² = 169 , 15² = 225 , 17² = 289 , 19² = 361.

Example 10: 121 , 484 , 1089 , 1936 , ? ,  4356
Answer : 11² = 121 , 22² = 484 , 33² = 1089 , 44² = 1936 , 55² = 3025 , 66² = 4356.

Example 11: 961 , 1024 , 1089 , ? 1225
Answer : 31² , 32² , 33² , 34² , 35²

Example 12: 1849 , ? , 2025 , 2116 , 2209
Answer : 43² , 44² , 45² , 46² , 47²

Example 13 : 2500 , 2401 , 2304 , ? , 2116 , 2025
Answer : 50² , 49² , 48² , 47² , 46² , 45²

Perfect Cube Series
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Perfect cube series is a arrangement of numbers in a certain order,where some numbers this Types of Series are based on cube of a number which is in same order and one cube number is missing in that given series.
we need to observe and find the accurate number to the series of numbers. This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam.Under below  given some more example for your better practice.
All numbers are arranged in sequence order. we need to observe and find the accurate number to this type series of numbers. Here we learn the perfect cube series of Example.
This type of problem are given in Quantitative Aptitude which is a very essential in banking exam. Under below  given some more example for your better practice.
In perfect cube series number is a combination of cube number are arranged. In example 1) 1331, 1728, 2197, ?  where you need to count them in a one step or two step calculation for obtain the difference common result according with the series of ratio numbers .
At first you can calculate missing number in ratio series  and  that you place the actual missing number in the ? or missing place. Be prepared when you calculate differences because it is either one or two step calculation. So when you calculate and get result of  two difference numbers you need follow some step wise.
At first calculate the first number cube value and second number cube value if all number are maintain a sequential order cube value then follow same steps which is carry up to last and after that you get actual missing number by finding the common value when you put the missing number you have noticed that all series numbers are common difference in between them.
This kind of missing series calculation you go thorough some common calculation shortcut tricks  using cube and cube shortcut tricks, or you memorize the 1 to 30 cube series number value.
In this type series example questions, it is sounds hard, but it really isn’t.  Get it? Once you have done this, by practice with more example then you just easily can do in your way as well competitive and as in bank exam also . So, each of our examples are given below.
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Perfect Cube Series:

Example 1: 3375, ?, 24389, 46656, 79507
Answer : 15³, 22³, 29³, 36³, 43³
(Each cube digit added with seven to become next cube number)

Example 2: 729, 6859, 24389, ?, 117649, 205379
Answer : 9³, 19³, 29³, 39³, 49³, 59³

Example 3: 1000, 8000, 27000, 64000, ?
Answer: 10³, 20³, 30³, 40³, 50³

Example 4: 1331 , ? , 35937 , 85184 , 166375
Answer : 11³ ,  22³ ,  33³ ,  44³ ,  55³
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Example 5: 125, ?, 343, 512, 729, 1000
Answer : 125 = 5³ , 216 = 6³, 343 = 7³, 512 = 8³, 729 = 9³, 1000 = 10³.

Example 6: 1 , 9 , 125 , 343 , ? , 729
Answer : 1³ , 3³ , 5³ , 7³ , 8³ , 9³

Example 7: 125, ?, 343, 512, 729, 1000
Answer: 125 = 5³, 216 = 6³, 343 = 7³, 512 = 8³, 729 = 9³, 1000 = 10³.

Example 8: 8 , 64 , ? , 512 , 1000 , 1728
Answer : 2³ , 4³ , 6³ , 8³ , 10³ , 12³

Example 9: 4096, 4913, 5832, ?, 8000
Answer: 4096 = 16³, 4913 = 17³, 5832 = 18³, 6859 = 19³, 8000 = 20³.
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Example 10: 1331 , ? ,  29791 , 68921 132651
Answer : 11³ , 21³ , 31³ , 41³ , 51³

Example 11: 1331, 1728, 2197, ?
Answer: 1331 = 11³, 1728 = 12³, 2197 = 13³, 2744 = 14³.
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Example  12: 1728, 1331, ?, 729, 512
Answer: 1728 = 12³, 1331 = 11³, 1000 = 10³, 729 = 9³, 512 = 8³.

Example  13: 1000 , 8000 , ? ,64000 , 125000
Answer : 10³ , 20³ , 30³ , 40³ , 50³

Example  14: 125000 , 64000 , ? , 8000 , 1000
Answer : 50³ , 40³ , 30³ , 20³ , 10³

Mixed Series

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Mixed Number series is a arrangement of numbers in a certain order. How you know that the given series is mixed series, notice that this  type of series are more then one different order which arranged in alternatively in a single series or created according to any non conventional rule.
Find the accurate number to the blank or ? mark series of numbers using calculation. This type of problem are given in Quantitative Aptitude which is a very essential  in banking exam.Under below  given some more example for your better practice.
In mixed Series a mixed number is a combination of number in another way it is not a sequential number series number that you have arranged. In example 1, 111, 220, 438, ?, 1746  where you need to count them in a one step or two step calculation for obtain the difference common result according with the series of mixed numbers .
At first you can calculate missing number in mixed series  and  that you place the actual missing number in the ? or missing place. Be prepared when you calculate differences because it is either one or two step calculation. So when you calculate and get two difference numbers result you need follow some step wise.
At first calculate the first and second number common difference then follow same steps  another two number differences calculation which is carry up to last and after that you get actual missing number by finding the common difference when you put the missing number you have noticed that all series number are common difference in between them.
This kind of missing series calculation you go through some common calculation shortcut tricks  square or division, cube, addition, multiplication.
In this type series example questions, it is sounds hard, but it really isn’t.  Get it? Once you have done this, by practice with more example then you just easily can do in your way as well competitive and as in bank exam also. So, each of our examples are given below.

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Example 1: 180, 179,183, 156, 172, ?
Answer : – 1³, + 2³, -3³, +4³, -5³

Example 2: 6, ?, 33, 69, 141, 285
Answer :  x 2 + 3, x 2 + 3, x 2 + 3, x 2 + 3, x 2 + 3, x 2 + 3

Example 3:
4, 16, 64, 256, 1024, ?
Answer: Multiply each number by 4 to get the next number.
4 x 4 = 16
16 x 4 = 64
64 x 4 = 256
256 x 4 = 1024
1024 x 4 = 4096

Example 4:
8, 16, 24, 40, 64, ?
Answer:
8 + 8 = 16
16 + 8( add previous ) = 24
24 + 16( add previous ) = 40
40 + 24( add previous ) = 64
64 + 40( add previous ) = 104

Examples 5:
24, ?, 208, 622, 1864
Answer:
from 24 to ? we get using this 24 x 3 = 72 – 2 = 70, Similarly we follow next steps
from 70 to 208 we get using this 70 x 3 = 210 – 2 = 208,
from 208 to 622 we get using this 208 x 3 = 624 – 2= 622,
from 622 to 1864 we get using this 622 x 3 = 1866 – 2 = 1864.
So the missing number is 70

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Examples 6:
111, 220, 438, ?, 1746
Answer:
from 111 to 220 we get using this 111 x 2 = 222 – 2 = 220,similarly we follow next steps
from 220 to 438 we get using this 220 x 2 = 440 – 2 = 438,
from 438 to ? we get using this 438 x 2 = 876 – 2 = 874,
from 874 to 1746 we get using this 874 x 2 = 1748 – 2 = 1746.
So the missing number is 874

Examples 7:
11, 24, 50, 102, 206, ?
Answer:
11 x 2 = 22 +2 = 24,
24 x 2 = 48 + 2 = 50,
50 x 2 = 100 + 2 = 102,
102 x 2 = 204 + 2 = 206,
206 x 2 = 412 + 2 = 414.
So the missing number is 414.

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Example 8:
0, 6, 24, 60, 120, 210, ?
Answer :
The given series is : 1³ – 1, 2³ – 2, 3³ – 3, 4³ – 4, 5³ – 5, 6³ – 6,
So the missing term = 7³ – 7 = 343 – 7 = 336 .

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Example 9:
11, 14, 19, 22, 27, 30, ?
Answer :
The pattern is + 3, + 5, + 3, + 5, …………
So the missing term is = 30 + 5 = 35 .

Example 10:
6, 12, 21, ? , 48
Answer :
The pattern is + 6, + 9, + 12, +15 ………..
So the missing term is = 21 + 12 = 33 .

Example 11:
18, 22, 30, ? ,78, 142
Answer :
The pattern is +4, +8, +16, +32, +64
So the missing term is = 30 + 16 = 46 .

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Example 12:
589245773, 89245773, 8924577, 924577, ?
Answer :
The pattern is The digits are removed one by one from the beginning and the end in order alternately, So to obtain the subsequent terms of the missing series is = 92457 .

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Example 13:
8, 35, ? , 143, 224, 323
Answer :
The pattern is (3² – 1), (6² – 1),………., (12² – 1), (15² – 1), (18² – 1)
So the missing term is = (9² – 1 ) = 81 – 1 = 80 .

Example 14:
3, 7, 23, 95, ?
Answer :
The pattern is ( x 2 + 1 ),( x 3 + 2) , ( x 4 + 3 ) , ……….
So the missing term is = 95 x 5 + 4 = 479 .

Geometric Series:

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Geometric Number series is a arrangement of numbers in a certain order, where some numbers are this type of series are based on ascending or descending order of numbers and each continues number is obtain by multiplication or division of the previous number with a static number.
we need to observe and find the accurate number to the series of numbers. This type of problem are given in Quantitative Aptitude which is a very essential in banking exam. Under below given some more example for your better practice.
In geometric series number is a combination of number arranged. In example 1) 5, 45, 405, 3645, ? where you need to count them in a one step or two step calculation for obtain the difference common result according with the series of numbers.
At first you can calculate missing number in geometric series and that you place the actual missing number in the ? or missing place. Be prepared when you calculate differences because it is either one or two step calculation. So when you calculate and get result of  two difference numbers you need follow some step wise.
At first calculate the first number 5 with 9 and second number value we get that is 45 then again second number calculate multiply with 9 and get the third number and follow same steps which is carry up to last and after that you get actual missing number by finding the common value when you put the missing number you have noticed that all series numbers are common 9 which multiply with number and get next number difference.
This kind of missing series calculation you go thorough some common calculation shortcut tricks  using cube multiplication division addition cube square shortcut tricks.
In this type series example questions, it is sounds hard, but it really isn’t.  Get it? Once you have done this, by practice with more examples then you just easily can do in your way as well competitive and as in bank exam also. So, each of our examples are given below.


Example 1: 3, ?, 21, 51, 162,
Answer : 3 x 0 +9 = 12, 12 x 1 + 9 = 21, 21 x 2 + 9 = 51, 51 x 3 + 9 = 162

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Examples 2: 5, 45, 405, 3645, ?
Answer: 5 x 9 = 45, 45 x 9 = 405, 405 x 9 = 3645, 3645 x 9 = 32805.

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Examples 3: 73205, 6655, 605, 55, ?
Answer: 5 x 11 = 55, 55 x 11 = 605, 605 x 11 = 6655, 6655 x 11 = 73205.

Example 4: 21, 84, 336, ?, 5376
Answer: 21 x 4 = 84, 84 x 4 = 336, 336 x 4 = 1344, 1344 x 4 = 5376

Example 5: 9, 54, ?, 1944, 11664
Answer: 9 x 6 = 54, 54 x 6 = 324, 324 x 6 = 1944, 1944 x 6 = 11664.

Example 6: 5, 35, ?, 1715, 12005,
Answer: 5 x 7 = 35, 35 x 7 = 245, 245 x 7 = 1715, 1715 x 7 = 12005.

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Example 7: 25, 100, ?, 1600, 6400
Answer: 25 x 4 = 100, 100 x 4 = 400, 400 x 4 = 1600, 1600 x 4 = 6400.

Example 8: 6, 18, 54, ?, 486, 1458
Answer: 6 x 3 = 18, 18 x 3 = 54, 54 x 3 = 162, 162 x 3 = 486, 486 x 3 = 1458.

Example 9: 15, 30, 60, 120, ?
Answer: 15 x 2 = 30, 30 x 2 = 60, 60 x 2 = 120, 120 x 2 = 240

Example 10: 19, ?, 475, 2375, 11875
Answer: 19 x 5 = 95, 475 x 5 = 2375, 2375×5 = 11875,

Example 11: 3, 48, ?, 768, 3072
Answer: 3 x 4 = 12, 12 x 4 = 48, 48 x 4 = 192, 192 x 4 = 768, 768 x 4 = 3072
 
Two stage Type Series:

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A two stage Arithmetic series is one in which the formation of arithmetic series are obtain from differences of continuous numbers themselves.
Example 1: i. 3, 9, 18, 35, 58,——
ii. 6, 9, 17, 23,———-

Number Series questions:

 

1) In each of the following questions a number series is given with one term missing. Choose the correct alternative that will continue the same pattern and fill in the blank spaces.

2, 7, 14, 23, ?, 47

 A.    31

 B.    28

 C.    34

 D.    38

Answer : C.

The given sequence is +5, +7, +9, ——

ie. 2+ 5 = 7, 7 + 7 = 14, 14 + 9 = 23

Missing Number = 23 + 11 = 34.

2) 4, 6, 12, 14, 28, 30, ?

 A.    32

 B.    64

 C.    62

 D.    60

Answer : D.

The given sequence is a combination of two series 4, 12, 28, .... and 6, 14, 30, .... The pattern is +8, +16, +32. So, the missing number = (28 + 32) = 60

3) 4, 9, 13, 22, 35, ?

 A.    57

 B.    70

 C.    63

 D.    75

Answer : A.

Sum of two consecutive numbers of the series gives the next number.

4) 11, 13, 17, 19, 23, 29, 31, 37, 41, ?

 A.    43

 B.    47

 C.    51

 D.    53

Answer : A.

The series consists of prime numbers.

5) 15, 31, 63, 127, 255, ?

 A.    513

 B.    511

 C.    523

 D.    517

Answer : B.

Each number is double of the preceding one plus 1.

6) 5, 11, 17, 25, 33, 43, ?

 A.    49

 B.    51

 C.    52

 D.    53

Answer : D.

The sequence is +6, +6, +8, +8, +10, ....

7) 9, 12, 11, 14, 13, ?, 15

 A.    12

 B.    16

 C.    10

 D.    17

Answer : B.

Alternatively, 3 is added and one is subtracted.

8) 0.5, 0.55, 0.65, 0.8, ?

 A.    0.7

 B.    0.9

 C.    0.95

 D.    1

Answer : D.

The pattern is + 0.05, + 0.10, + 0.15, .....

9) 1, 4, 9, 16, 25, ?

 A.    35

 B.    36

 C.    48

 D.    49

Answer : B.

The sequence is a series of squares, 12, 22, 32, 42, 52....

10) 2, 1, (1/2), (1/4), ?

 A.    (1/3)

 B.    (1/8)

 C.    (2/8)

 D.    (1/16)

Answer : B.

This is a simple division series; each number is one-half of the previous number.

In each of the following questions a number series is given with one term missing. Choose the correct alternative that will continue the same pattern and fill in the blank spaces.

11) 1, 4, 27, 16, ?, 36, 343

 A.    125

 B.    50

 C.    78

 D.    132

Answer : A.

The series consists of cubes of odd numbers and square of even numbers.

12) 20, 19, 17, ?, 10, 5

 A.    15

 B.    14

 C.    13

 D.    12

Answer : B.

The Pattern is - 1, - 2, -3, ...

13) 7, 10, 8, 11, 9, 12, ?

 A.    13

 B.    12

 C.    10

 D.    7

Answer : C.

This is a simple alternating addition and subtraction series. In the first pattern, 3 is added; in the second, 2 is subtracted.

14) 6, 11, 21, 36, 56, ?

 A.    51

 B.    71

 C.    81

 D.    41

Answer : C.

The pattern is + 5, + 10, + 15, + 20,....

15) 2, 3, 5, 7, 11, ?, 17

 A.    15

 B.    14

 C.    13

 D.    12

Answer : C.

The series consists of prime numbers starting from 2.

16) 36, 34, 30, 28, 24, ?

 A.    26

 B.    23

 C.    22

 D.    20

Answer : C.

This is an alternating number subtraction series. The pattern is -2, -4, -2, ....

17) 13, 35, 57, 79, 911, ?

 A.    1145

 B.    1113

 C.    1117

 D.    1110

Answer : B.

The terms are formed by joining together consecutive odd numbers in order. i.e. 1 and 3, 3 and 5, 5 and 7, 7 and 9, 9 and 11,....

18) 6, 11, 21, 36, 56, ?

 A.    65

 B.    78

 C.    81

 D.    97

Answer : C.

The pattern is + 5, + 10, + 15, + 20,...

19) 53, 53, 40, 40, 27, 27, ?

 A.    14

 B.    12

 C.    16

 D.    18

Answer : A.

First, each number is repeated, then 13 is subtracted to arrive at the next number.

20) 11, 10, ?, 100, 1001, 1000, 10001

 A.    1000

 B.    121

 C.    111

 D.    101

Answer : D.

The pattern is - 1, × 10 + 1, - 1, × 10 + 1, - 1, × 10 + 1, ....

In each of the following questions a number series is given with one term missing. Choose the correct alternative that will continue the same pattern and fill in the blank spaces.

21) 1, 6, 13, 22, 33, ?

 A.    35

 B.    46

 C.    38

 D.    49

Answer : B.

The pattern is + 5, + 7, + 9, + 11,....

22) 21, 9, 21, 11, 21, 13, 21, ?

 A.    15

 B.    17

 C.    23

 D.    25

Answer : A.

In this alternating repetition series, the number 21 is interpolated. If you exclude 21, the series increases by 2, beginning with the number 9.

23) 10, 14, 26, 42, 70, ?

 A.    86

 B.    98

 C.    114

 D.    126

Answer : C.

Each term of the series, except the first two terms, is 2 more than the sum of the preceding two terms.

24) 1, 9, 17, 33, 49, 73, ?

 A.    78

 B.    85

 C.    91

 D.    97

Answer : D.

The pattern is + 8, + 8, + 16, + 24,...

25) 31, 29, 24, 22, 17, ?

 A.    15

 B.    23

 C.    13

 D.    25

Answer : A.

This is a simple alternating subtraction series, with a pattern -2, -5, -2, -5 ....

26) 1, 9, 25, 49, 81, ?

 A.    100

 B.    121

 C.    144

 D.    169

Answer : B.

The series consists of squares of consecutive prime numbers.

27) 5, 9, 17, 29, 45, ?

 A.    65

 B.    56

 C.    74

 D.    57

Answer : A.

The pattern is + 4, + 8, + 12, + 16, ....

28) 14, 28, 20, 40, 32, 64, ?

 A.    56

 B.    46

 C.    58

 D.    48

Answer : A.

This is an alternating multiplication and subtracting series with a pattern x2, -8, x2, -8.

29) 2, 6, 12, 20, 30, 42, 56, ?

 A.    63

 B.    67

 C.    69

 D.    72

Answer : D.

The pattern is 1 × 2, 2 × 3, 3 × 4, 4 × 5, 6 × 7, 7 × 8,....

30) 3, 7, 15, 31, 63, ?

 A.    89

 B.    127

 C.    142

 D.    158

Answer : B.

Each number in the series is the preceding number multiplied by 2 and then increased by 1.

अब आप निम्न प्रश्नों को हल करें, उत्तर 1a,2b,3c,4d,5e,6b प्रारूप में Comments Section में पोस्ट करें. आपको प्रत्येक प्रश्न का विस्तृत उत्तर भेजा जायेगा. पूरे प्रश्नपत्र को व ऐसे ही अन्य प्रश्नपत्रों को प्राप्त करने / डाउनलोड के लिये कमेंट बाक्स में अपना ईमेल ऐड्रेस, वाट्सऐप नंबर लिखें या https://www.facebook.com/groups/NextGenCareers/से डाउनलोड कर सकते है>

TEST No:NGC-163-QA-LSSC-E049 :: No of Questions:049   ::   Time Allowed : 39  Minutes
Directions: Please Tick Appropiate Option and Submit, Inform Serial Number of Questions where you need detailed Explanations to bhagirathprayash@gmail.com or Massage to 9462900411

Question 1:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Number Series

Prime Number Series:

Example 1. 4, 9, 25, 49, 121, 169,…
(a) 324  (b) 289  (c) 225   (d) 196

Answer: (b) 289
Solution. (b) The given series is a consecutive square of prime number series. The next prime number is 289.

Question 2:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 2. 5, 7, 13, 23, …
(a) 25   (b) 27  (c) 29   (d) 41

Answer: (d) 41
Solution. (d) The difference between prime numbers is increasing. 7 is next prime to 5; 13 is second to next prime to 7; 23 is third to next to 13. Hence, next should be fourth to next prime to 23. Hence, required number is 41.

Question 3:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Multiplication Series:

Example 3. 4, 8, 16, 32, 64… 256
(a) 96   (b) 98  (c) 86  (d) 106

Answer: (a) 96
Solution. (a) The numbers are multiplied by 2 to get the next number.
64 × 2 = 128

Question 4:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 4. 5, 20, 80, 320, … 1280
(a) 5120  (b) 5220  (c) 4860  (d) 3642

Answer: (a) 5120
Solution. (a) The numbers are multiplied by 4 to get the next number.
1280 × 4 = 5120

Question 5:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Difference Series:

Example 5. 3,6,9,12,15,…. 21
(a) 16   (b) 17  (c) 20   (d) 18

Answer: (d) 18
Solution. (d) The difference between the numbers is 3.
15 + 3 = 18

Question 6:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 6. 55, 50, 45, 40,….30
(a) 33   (b) 34  (c) 35   (d) 36

Answer: (c) 35
Solution. (c) The difference between the numbers is -5.
40 – 5 = 35

Question 7:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Division Series:

Example 7. 5040, 720, 120, 24, ….2,1
(a) 8   (b) 7  (c) 6   (d) 5

Answer: (c) 6
Solution. (c)

Question 8:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 8. 16, 24, 36,… 81
(a) 52   (b) 54  (c) 56   (d) 58

Answer: (b) 54
Solution. (b) Previous number × = Next number

Question 9:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

n2 Series

Example 9. 4, 16, 36, 64, …. 144
(a) 112   (b) 78  (c) 100   (d) 81

Answer: (c) 100
Solution. (c) The series is square of consecutive even numbers. 22, 42,62, 82
Next number is 102 = 100

Question 10:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 10. 1, 4, 9, 16, 25, 36, 49, … 81
(a) 100   (b) 121  (c) 64   (d) 144

Answer: (c) 64
Solution. (c) The series is 12, 22, 32, 42, 52,62, 72,….
The next number is 82 = 64

Question 11:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

(n2 + 1) Series

Example 11. 17, 26, 37, 50, 65,….101
(a) 82   (b) 75  (c) 78   (d) 90

Answer: (a) 82
Solution. (a) The series is 42 + 1, 52 +1, 62 + 1, 72 + 1, 82 + 1.
The next number is 92 + 1 = 82

Question 12:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 12. 101, 401, 901, 1601, 2501, …. 4901
(a) 2201  (b) 3301  (c) 4401   (d) 3601

Answer: (d) 3601
Solution. (d) The series is 102 + 1, 202 +1, 302 + 1, 402 + 1, 502 + 1, etc.
The next number is 602 + 1 = 3601

Question 13:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

(n2 -1) Series

Example 13. 3, 8, 15, 24,…48
(a) 32  (b) 33  (c) 34   (d) 35

Answer: (d) 35
Solution. (d) The series is 22 – 1, 32 –1, 42 – 1,52 – 1. etc.
The next number is 62 – 1 =35

Question 14:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 14. 99, 80, 63,….35
(a) 48   (b) 84  (c) 46   (d) 64

Answer: (a) 48
Solution. (a) The series is 102 -1, 92 -1, 82 -1, etc.
The next number is 72 – 1 = 48

Question 15:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

(n2 + n) Series

Example 15. 2, 6, 12, 20, 30,…. 56
(a) 32  (b) 34  (c) 42  (d) 24

Answer: (c) 42
Solution. (c) The series is 12 + 1, 22 + 2, 32 + 3, 42 + 4, 52 + 5, etc.
The next number is 62 + 6 = 42

Question 16:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 16. 110, 132, 156, 182,….
(a) 212   (b) 201  (c) 211   (d) 210

Answer: (d) 210
Solution. (d) The series is 102 + 10, 112 + 11, 122 + 12, etc.
The next number is 142 + 14 = 210

Question 17:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

(n2 – n) Series

Example 17. 0, 2, 6, 12, 20,….42
(a) 25   (b) 30  (c) 32   (d) 40

Answer: (b) 30
Solution. (b) The series is 12 – 1 = 0, 22 – 2 = 2, 32 – 3 = 6, etc.
The next number is 62 – 6 = 30

Question 18:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 18. 90, 380, 870, 1560,…..
(a) 2405   (b) 2450  (a) 2400  (d) 2455

Answer: 2
Solution. (b) The series is 102 – 10, 202 – 20, 302 – 30, etc.
The next number is 502 – 50 = 2450

Question 19:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

n3 Series

Example 19. 1, 8, 27, 64,…. 216
(a) 125   (b) 512  (c) 215   (d) 122

Answer: (a) 125
Solution. (a) The series is 13, 23, 33 , 43, etc.
The next number is 53 = 125

Question 20:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 20. 1000, 8000, 27000, 64000,….
(a) 21600   (b) 125000  (c) 152000   (d) 261000

Answer: (b) 125000
Solution. (b) The series is 103 , 203, 303, 403, etc.
The next number is 503 = 125000

Question 21:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

(n3 + 1) Series

Example 21. 2, 9, 28, 65,…217
(a) 123   (b) 124  (c) 125   (d) 126

Answer: (d) 126
Solution. (d) The series is 13 +1, 23 + 1, 33 + 1, etc.
The next number is 53 + 1 = 126

Question 22:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 22. 1001, 8001, 27001, 64001, 125001,….
(a) 261001   (b) 216001  (c) 200116   (d) 210016

Answer: (b) 216001
Solution. (b) The series is 103 + 1, 203 + 1, 303 + 1, etc.
The next number is 603 + 1 = 216001

Question 23:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

(n3 -1) Series

Example 23. 0, 7, 26, 63, 124,…
(a) 251   (b) 125  (c) 215   (d) 512

Answer: (c) 215
Solution. (c) The series is 13 – 1, 23 – 1, 33 – 1, etc.
The next number is 63 – 1 = 215

Question 24:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 24. 999, 7999, 26999, 63999,….
(a) 199924   (b) 124999  (c) 129994   (d) 999124

Answer: (b) 124999
Solution. (b) The series is 103 – 1, 203 – 1, 303 – 1, etc.
The next number is 503 – 1 = 124999

Question 25:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

(n3 + n) Series

Example 25. 2, 10, 30, 68,….222
(a) 130  (b) 120  (c) 110   (d) 100

Answer: (a) 130
Solution. (a) The series is 13 + 1, 23 + 2, 33 + 3, etc.
The next number is 53 + 5 = 130

Question 26:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 26. 1010, 8020, 27030, 64040,….
(a) 125500   (b) 125050  (c) 100255  (d) 120055

Answer: (b) 125050
Solution. (b) The series is 103 + 10 = 1010, 203 + 20 = 8020, etc.
The next number is 503 + 50 = 125050

Question 27:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

(n3 – n) Series
Example 27. 0, 6, 24, 60,…. 210
(a) 012   (b) 210  (c) 201   (d) 120

Answer: (d) 120
Solution. (d) The series is 13 – 1 = 0, 23 – 2 = 6, 33 – 3 = 24, etc.
The next number is 53 – 5 = 120

Question 28:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 28. 990, 7980, 26970, 63960,….
(a) 124500   (b) 124005  (c) 120045   (d) 124950

Answer: (d) 124950
Solution. (d) The series is 103 – 10, 203 – 20, 303 – 30 etc.
The next number is 503 – 50 = 124950

Question 29:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Letter Series - Type 1

One Letter Series Such series consists of one letter in each term and this series is based on increasing or decreasing positions of corresponding letters according to English alphabet.
Example 1: B, C, A, D, Z, E, … F, X, G
(a) U   (b) Y  (c) W   (d) V

Answer: (b) Y
Solution. (b) The sequence consists of two series B, A, Z, Y, X and C, D, E, F, G. The missing letter is Y.

Question 30:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 2: P, U, Z, … J, 0, T
(a) E  (b) U  (c) S   (d) P

Answer: (a) E
Solution. (a) The sequence is P+ 5, U+ 5,Z+ 5. The missing letter is Z + 5 = E

Question 31:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 3: B, D, G, I, … N, Q, S
(a) I   (b) J  (c) L   (d) K

Answer: (c) L
Solution. (c) The sequence is B + 2, D+ 3, G + 2, I + 3 and so on.

Question 32:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Letter Series - Type 12

Two Letter Series The first letters of the series follow one logic and the second letters follow another logic.
Example 4: EZ, DX, CV,..., AR, ZP
(a) CS   (b) AM  (c) BT   (d) TG

Answer: (c) BT
Solution. (c) First and second letters follow a sequence of-1 and -2 respectively.

Question 33:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 5: DG, HK, LO, PS, TW,…
(a) XA   (b) ZA  (c) XB   (d) None of these

Answer: (a) XA
Solution. (a) First and second letters follow a sequence of + 4.

Question 34:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 6: DX, EY FV, ... : ; HT, IU
(a) HV   (b) IX   (c) GW   (d) BZ

Answer: (c) GW
Solution. (c) First, -third and fifth terms follow a sequencee and second, fourth and sixth terms follow another sequence. (DX, FV, HT, etc) and (EY, GW, IU, etc).

Question 35:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Letter Series - Type 3

Three Letter Series: :Such series consist of three letters in each term. The first letters follow one logic, the second letters follow another logic and the third letters follow some other logic.
Example 7: DIE, XCY, RWS, ...
(a) LQN   (b) QMP  (c) LMS   (d) LQM

Answer: (d) LQM
Solution. (d) First, second and third letters of each group follow a sequence of -6 series.

Question 36:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 8: VPG, UQF, ..., SSD, RTC
(a) SQD   (b) TRE  (c) TRS   (d) QDT

Answer: (b) TRE
Solution. (b) First, second and third letters follow a sequence of –1, + 1, –1 series respectively.

Question 37:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Example 9: DJS, HNW, LRA, PVE, ..., XDM
(a) TZI   (b) SAF  (c) UXH   (d) None of these

Answer: (a) TZI
Solution. (a) First, second and third letters follow a sequence of + 4 series.

Question 38:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Letter Series - Type 4

A series of letters is given with one or more missing letters. From the choices, the choice that gives the letters that go into the blanks has to be selected as answer.
Example 10: In the following series some letters are missing. From the choices, select the choice that gives that letters that can fill the blanks in the given sequence.
a_ c_ b_ab_a_ca_c
(a) abaccb  (b) accbab  (c) aabbcc   (d) baccbb

Answer: (d) baccbb
Solution. (d) First of all, notice that there are 6 blanks in the given sequence and each choice gives six letters to fill the six blank in order. Now, we have to select an alternative which if placed in the blanks of the series in order, we get a complete series of letters which follow some particular pattern.
The best way is to try with each option. Inserting the letters of option (d) in place of the blanks, we get a series like “abc abc abc abc abc” which is a repetition of the group of letters “abc”.

Question 39:  NEXTGEN CAREER COACHING#9462900411--NGC-163-QA  

Letter Series - Type 5

Here, students are asked to count how many times a particular letter or group of letters satisfying some conditions occurs and mark that number as the answer choice.
Example 11: In the following sequence of letters, in how many instances the letters n is immediately preceded by the letter t ?
s n r u a t n n g h j t k n s t n d g c l n t t t n n n t n t n t s m v b t n g c x d p t n k l s t n t
(a) 5   (b) 6  (c) 7  (d) 8

Answer: (d) 8
Solution. (d) On counting, we find that the letter n occurs 8 times, where n is immediately preceded by the letter t.

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