# Addition Tricks for Bank Exams – SBI Clerks and PO’s 2017

To crack SBI clerk’s exam we need to be smart in doing calculations faster and accurate. So we have to use some maths tricks while solving aptitude section that traditional method of solving. Here we are providing some simple addition tricks with examples which are very much useful to crack upcoming SBI exams 2016.

Some Tricky Additions – Simplifications

IF you came across these type of additions we can follow this easy trick to solve in less time.

**Model 1:**

** 1) 6 + 66 + 666+ 6666+ 66666+ 666666**

= 6 × (123456) -> * [Explanation: *In above problem the number 6 repeated orderly 1 time, 2 times …up to 6 times).]

**= 740736.**

**2) 4 + 44 + 444 + 4444 + 44444 + 444444**

= 4 × (123456)**= 493824.**

**3) 7 + 77 + 777 + 7777 + 77777**

= 7 × (1 2 3 4 5)**= 86415**

**Model 2:****1) 6.6 + 66.66 + 666.666 + 6666.6666 + 66666.66666**

= 6 × (12345) + [6 × (54321)]-> [**Explanation:** In above problem the number 6 repeated orderly 1 time, 2 times …up to 6 times).]= 74070 + 3.25926 ( 5 numbers after decimal point)**= 74073.25926**

**2) 4.4 + 44.44 + 444.444 + 4444.4444**

= [4× ( 1 2 3 4)] + [ 4×(4 3 2 1)]

= 4936 + 1.7284 ( 4 times repeatations after decimal point so keep decimal point upto 4 digits )

**= 4937.7284.**

**Model 3:**

### 1) 9.4 + 99.44 + 999.444 + 9999.4444

**Step 1:**

Before decimal point 9 repeated up to 4 times orderly from 1 time, 2 times …4 times. so multiply 9 with 1234.

[9 ×( 1234) ] = **11106****Step 2:**

After decimal point 4 repeated up to 4 times orderly from 1 to 4 numbers. So multiply 4 with (4321) and put the decimal point after 4 digits by counting from right to left.

[4 × (4321) ] = **1. 7284****Step 3:**

Add step 1 + step2

11106 + 1.7284 = **11107.7284****Model 4:****What is the sum of all even natural numbers below 100****Solution:**

2 + 4+ 6 + 8 + 10 +……+98

Formula: Sum =n/2 (f + l) [n = total numbers, f = first number, l= last number].

= 49/2 ( 2 + 98) = 49 × 50 =** 2450**

**Examples: **

**1. 2546 + 5480 + 3210**

Here add all first digits in step 1, and then add all second digits and then 3^{rd} digits and so on as shown below.

2546 + 5480 + 3210

10000 + 1100 + 130 + 6 = ** 11, 236.**

**2. 256 + 516 + 310.**

1000 + 70 + 12 = 1082.

**3. 5467 + 6752 + 8971 +8605.**

**4. 679 + 760 + 896**

Add all first digits in step 1, and then add all second digits and then 3^{rd} digits and so on as shown below.

2100 + 220 + 15.

2335.

**5.6587 + 6540 + 5555 + 2345.**

19000 + 1800 + 210 + 17

add from right to left,

21, 027

**IBPS Shortcut Formulas useful to solve simplifications faster.**

**(a**= (a-b) (a^{3 }– b^{3})^{2}+ ab + b^{2})**(a**= (a+b) (a^{3 }+ b^{3})^{2}– ab + b^{2})**(a + b)**= 2( a^{2}+ (a – b)^{2}^{2}+ b^{2})**a**= (a –b) (a+b) (a^{4}– b^{4}^{2}– b^{2})**(a**=(a-b) [a^{5}-b^{5})^{4}+a^{3}b+a^{2}b^{2}+ab^{3}+b^{4}].

**Sum of first n odd numbers**

= 1 + 3 + 5 + ………… + (2n-1)= **n ^{2}**

**Sum of first n even numbers**

= 2 + 4 + 6 + …………+ 2n =** n (n+1).**

**Sum of first n natural numbers**

= 1+2+3 +4 + ……..+n = **n(n+1)/2.**

**Sum of squares of first n natural numbers**

=1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2} + …………+n^{2}.

= **[n (n+1)(2n+1)]/6.**

**Sum of cubes of first n natural numbers**

= 1^{3} + 2^{3}+3^{3} + 4^{3} + …… + n^{3}

= **[n(n+1)/2] ^{2}**

**1 + 2/3 + (2/3) ^{2} + (2/3)^{3} + ……(2/3)^{n}**=

**1/(1 -2/3)**

^{n }**1/1.3 + 1/3.5 + 1/5.7 +.……. + 1/[(2n+1) (2n-1)]** =?

**Shortcut method:**

1/gap [1/first – 1/(2n-1)]= n/(2n+1) last number = 2n-1 =7

= 2n=8

=n=4.

= 4/(2*4 + 1) = 4/9

**Geometric Progression:** When the sum looks likes as below we have to find the total sum by using formula.

a + ar^{2} + ar^{3} + ar^{4 }+ …………… + ar^{n}

^{ }= Total sum = a (1-r^{n})/ 1-r. [ a = first number]

Example: 2 + 2×3^{2} + 2×3^{3} + 2×3^{4} + 2×3^{5}

= 2( 1- 3^{5})/1-3

= 2( 1- 243)/(-2)

=242

**Athematic Progression:**

The sum of a, (a+d), (a+2d), (a+3d),……… (a+(n-1)d)

Total sum = n/2 [2a + (n-1) d]

= **n/2 (first +last).**

**What is the sum of all even natural numbers below 100**

2 + 4+ 6 + 8 + 10 +……+98

**Formula:**

Sum =n/2 (f + l) [n = total numbers, f = first number, l= last number].

= 49/2 ( 2 + 98) = 49 × 50 = **2450**

Here, we can split 12 as 4×3

Ans. is 4 .

Here, we can split 12 as 4 × 3.

we take small number 3 as answer.

Formula:

A^{(2 ^n -1)/2^n)}

3^{(2 ^4 -1)/2^4)}

= 3^{(15/16)}

= 6

sqrt| 4 + 5/16 + 6/25 + 7/36 ( solve from right to left)

Explaination: ( sqrt (49) = 7, then 7+ 29 =36, sqrt(36) =6, 6+19 =25, sqrt (25) =5…..)

sqrt(4 +5) = sqrt(9) = 3