Problems On Ages Shortcuts

Problems-on-Ages-Shortcuts

Problems on Ages is one of the questions asked in every Bank Exam. When you know that this is a sure question, why not prepare yourself well to stand up to the competition. Equip yourself with this Problems on Ages shortcuts and you can solve problems on ages like a PRO. And if you know divisibility tricks, its cake walk for you. You can use both tricks to solve the problem quickly with accuracy.

Let me walk you through few examples to get an understanding of the shortcut.

Problems on Ages Examples

Example 1:
Present ages of S and A are in the ratio of 5:4 respectively. Three years hence, the ratio of their ages will become 11:9 respectively. What is A’s present age in years?
A)40 years
B)43 years
C)24 years
D)27 years
E)None of the above

Step 1:Write down the condition 1 given in question as follows
Present ages of S & A= 5:4

And we need to find A’s present age.

Step 2: Observe the options given in answer and find out which option is exactly divisible by 4.
We see that,

option A) 40/4=10 Exactly divisible by 4 so this could be the answer.
option B) 43/4 Not exactly divisible by 4
option C) 24/4=6 Exactly divisible by 4 so this could be the answer.
option D) 27/4 Not exactly divisible by 4.

So we have 2 options A and C exactly divisible by 4.

Step 3: Since we have 2 options satisfying condition 1. We need to check the options for condition 2 to see which among them is the answer.


Step 4: Condition 2 is Three years hence, the ratio is 11:9 is given.
First, we shall add 3 years to each of the options to get the age after 3 years(3 years hence)and divide each of the options by 9(A’s part)
We see that,

Option A) 40+3=43 years 43/9= Not exactly divisible by 9.
Option B) 43+3=46 years 46/9= Not exactly divisible by 9.
Option C) 24+3=27 years 27/9= Exactly divisible by 9.
Option B) 27+3=30 years 30/9= Not exactly divisible by 9.

We see that only option C satisfies condition 2.

Therefore, we can say that option C is the correct answer as it satisfies both condition 1 and condition 2.

A’s present age is 24 years.

Example 2:
Six years ago, the ratio of the ages of K and S was 6:5. Four years hence, the ratio of their ages will be 11:10. What is S’s age at present?
A)18 years
B)16 years
C)20 years
D)25 years
E)None of the above

Step 1:  Condition1 given in the question is 6 years ago, the ratio was 6:5 is given
First, we shall subtract 6 years from each of the options to get the age before 6 years(6 years ago)and divide each of the options by 5(S’s part)
We see that,

Option A) 18-6=12 years 12/5 Not exactly divisible by 5.
Option B) 16-6=10 years 10/5=5 Exactly divisible by 5.
Option C) 20-6=14 years 14/5 Not exactly divisible by 5.
Option D) 25-6=19 years 19/5 Not exactly divisible by 5.

We see that only option B satisfies condition 1.


Thus, we can say that option C is the correct answer as it is the only option which satisfies the condition1.
We do not have 2 options satisfying condition1, so no need to check if the options satisfy condition 2. We can stop here and save a few seconds here.

S’s present age is 16 years.

IBPS SO Interview Call Letter|Download

IBPS-SO-Interview-Call-letter

Institute of Banking Personnel Selection(IBPS) releases IBPS SO Interview Call letter today. You can download the Interview call letter from the official IBPS website www.ibps.in.

Important dates

Commencement of call letter download 14th Feb 2019

Closure of call letter download 2nd Mar 2019

How to download the SO Interview call letter?

  • Go to the link https://ibpsonline.ibps.in/crpspl8nov18/clinta_feb19/login.php?appid=ec2543ac27e639c6bf30028899164cf8
  • Enter your credentials such as Registration No or Roll No and password or your date of birth (dd-mm-yy) in the required places. Ensure that the date of birth is the same as the one provided at the time of registration.
  • Enter the captcha as shown in the form.
  • Click on Login and download the interview call letter.

Square Root Shortcut Nonperfect Squares

Square-Root-Shortcut-Nonperfect-Squares

In the previous trick, we learned how to find the square root of any number which is a perfect square. Now, using this square root shortcut, you can easily find the square root of a nonperfect square. For instance, lets say you want to find √26. Guess what, you can find the square root of this nonperfect square in seconds. Here is how you can do it.

Square Root Shortcut for Nonperfect squares steps

Step1: Find the nearest square less than or greater than a perfect square and find its square root.
Step 2: Now, find the difference between the nearest square and the given number.
Step 3: Multiply the result obtained in step 1 by 2
Step 4: Divide the result obtained in step 2 by the result obtained in step 3.
Step 5: Combine the results of step 1 and step 4 to arrive at your answer.

Remember the below squares
12=1
22=4
32=9
42=16
52=25
62=36
72=49
82=64
92=81
102=100
112=121
122=144
132=169
142=196
152=225
162=256
172=289
182=324
192=361
202=400

Let us now see a few examples to understand better

Examples

Example 1: Find √26

Step 1: Finding the nearest square to number 26. We find that 25 is the nearest square to 26. √25 =5

Step 2: Finding the difference between the nearest square and the given number. Here, 26-25= 1. Difference= 1

Step 3: Multiplying the result obtained in step 1 by 2.

We get, 5 X 2=10

Step 4: Dividing the result obtained in step 2 by result obtained in step 3.

We get, 1/10=0.10

Step 5: Combining the results of step 1 and step 4.

We get, 5 + 0.10= 5.10

Ans √26 = 5.10

Example 1: Find √38

Step 1: Finding the nearest square to number 38. We find that 36 is the nearest square to 38. √36 =6

Step 2: Finding the difference between the nearest square and the given number. Here, 38-36= 2. Difference= 2

Step 3: Multiplying the result obtained in step 1 by 2.

We get, 6 X 2=12

Step 4: Dividing the result obtained in step 2 by result obtained in step 3.

We get, 2/12=0.16

Step 5: Combining the results of step 1 and step 4.

We get, 6 + 0.16= 6.16

Ans √38 = 6.16

Also see Shortcut to find the square root of any perfect square

IBPS SO Result Out: CRP-SPL-VIII Main Exams

IBPS So Result Out

IBPS SO Result of the CRP-SPL-VIII main exams released today.
Institute of Banking Personnel Selection (IBPS) announces the main exams results held on 27th Jan 2019 for the recruitment of specialist officer.

Important dates

Commencement of result 6th Feb 2019

Closure of result 12th Feb 2019

Check your IBPS SO Result via official IBPS site below

https://ibpsonline.ibps.in/crpspl8nov18/res8stda_feb19/login.php?appid=544647cc4b08724bd1a0f29cb03256e4

Sum of Consecutive Numbers Shortcut

sum of consecutive numbers shortcut

Finding the sum of consecutive numbers is a common question asked in competitive exams. You can now find the sum of consecutive numbers starting with 1, sum of consecutive odd numbers and sum of consecutive
even numbers in seconds.

What are consecutive numbers?

Numbers that follow each other in order are Consecutive numbers.Here the difference between every two numbers is 1. The consecutive numbers can be represented as n,n+1,n+2.
Examples of consecutive numbers:
(1)1,2,3,4,5,6,7,8
(2)6,7,8,9,10,11

Sum of consecutive numbers starting from 1 shortcut

Now that we know what consecutive numbers are.We shall see a shortcut to find sum of consecutive numbers starting from 1.
It can be represented as Sum of n consecutive numbers = n+(n+1)+(n+2)

Sum of consecutive numbers starting from 1 shortcut steps

Step 1:Count the total number of integers from 1 to the last consecutive number given.
Step 2:Multiply the result in step 1 by 1 more than the number.
Step 3:Divide the result of step 2 by 2.

We shall see some examples to understand better
Example 1:Find the sum of all consecutive numbers from 1 to 9.
Step 1: Counting number of integers from 1 to 9. We find that there are 9 integers.

Step2: Multiplyig the number of integers(9) by 1 more than the number that is 10.
We get, 9 X 10=90

Step3: Dividing the result obtained in step2 by 2
we get, 90/2=45

Ans Hence, sum of consecutive numbers from 1 to 9=45

Example 2:Find the sum of all consecutive numbers from 1 to 99.
Step 1: Counting number of integers from 1 to 99. We find that there are 99 integers.

Step2: Multiplyig the number of integers(99) by 1 more than the number that is 100.
We get, 99 X 100=9900

Step3: Dividing the result obtained in step2 by 2
we get, 9900/2=4950

Ans Hence, sum of consecutive numbers from 1 to 99=4,950

Odd Consecutive numbers

Odd numbers that follow each other are consecutive odd numbers. Here the difference between every two odd numbers is 2.
It can be represented as Sum of n numbers = n+(n+2)+(n+4)+(n+6) where n is an odd number..
We shall see how to add odd consecutive numbers.

Sum of consecutive odd numbers starting from 1 steps

Step 1: Count the total number of odd integers from 1 to the last consecutive number given.
Step 2: Square the result obtained in step 1.

Example 1:Find the sum of all consecutive odd numbers from 1 to 10.
Step 1: Counting number of integers from 1 to 9. We find that there are 5 integers.
Step 2: Squaring the result obtained in step 1.
We get, 52=25
Ans hence, Sum of odd numbers from 1 to 9 is 25

Example 2:Find the sum of all consecutive odd numbers from 1 to 100.
Step 1: Counting number of integers from 1 to 100. We find that there are 50 integers.
Step 2: Squaring the result obtained in step 1.
We get, 502=2500
Ans 2500

Even Consecutive numbers

Even numbers that follow each other are consecutive even numbers. Here the difference between every two even numbers is 2.
It can be represented as Sum of n numbers = n+(n+2)+(n+4)+(n+6)where n is an even number.
We shall see how to add even consecutive numbers.

Sum of consecutive even numbers starting from 2 steps

Step 1: Count the total number of odd integers from 1 to the last consecutive number given.
Step 2: Multiply the result in step 1 by 1 more than the number.

Example 1:Find the sum of all consecutive even numbers from 1 to 10.

Step 1: Counting number of integers from 1 to 10. We find that there are 5 integers.
Step 2: Multiplying the number of integers(5) by 1 more than the number that is 6.
We get, 5 X 6=30
Ans 30

Example 2:Find the sum of all consecutive even numbers from 1 to 100.
Step 1: Counting number of integers from 1 to 100. We find that there are 50 integers.
Step 2: Multiplying the number of integers(50) by 1 more than the number that is 51.
We get, 50 X 51=2550
Ans 2550.