Numerical Ability is one of the heebie-jeebie topics which consuming more time in Banking exams. There is no doubtย Numerical Ability questions will be a complexย of mathematics. Once you got a right strategyย to solve the Numerical Ability questions then youย can easily manage your time. To get a maximum mark in this section the aspirants need to know the tricks of Numerical Ability. Don’tย worry if you are a basic learner of banking exam Read and understand the tricks ofย Numerical Ability. Here you can find the set of the question with important tricks to solveย Numerical Ability questions.

**Daily Practice Questions for Numerical Ability:**

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**Important Tricks to Solve Numerical Ability Questions:**

**Topics From Numerical Ability:**

- Shortcut Tricks to find the Square of Numbers
- Easy Method to Take Square Root for a Number
- Shortcut for the Multiplication of 2 Digit & 3 Digit Numbers
- Shortcut for the Multiplication of 3 Digit & 4 Digit Numbers
- Easy Method to Take Cube Root of Numbers
- Shortcut Rules for Basic Multiplication
- Shortcut Rules for Basic Subtraction
- Shortcut Rules for Basic Division

**Numerical Ability – Shortcut Tricks to find the Square of Numbers:**

**SQUARE OF NUMBERS:**

- Square of a number is the product obtained by multiplying a number by itself

2ร2 = 4

11ร11 = 121 - To find the square of two-digit numbers (10-99) we can consider the following steps

1) Let the 2-digit number be = AB

2) Now to find AB2

3) Unit digit of square = B2

4) Tenโs digit of square = 2รAรB (+ Carry if any from the previous step)

5) The rest of the digits of square = A2 (+ Carry if any from the previous step)

**FOR EXAMPLE**

67^{2} = ?

AB^{2} = 672

**STEP 1:**

B^{2} = 72 = 4** 9ย **Here โ9โ is the unit digit and โ4โ is carry

- 67
^{2}= _ _ _ 9

**STEP 2:**

[2รAรB] + Carry from previous step i.e., 4

2รAรB = 2ร6ร7 = 84

Add Carry โ4โ with the above โ84โ we get [84+4= 8**8**]

Therefore at the end of the 2nd Step

From Step 1, here โ8โ is the unit digit and โ8โ is carry

- ย 67
^{2}= _ _ 8 9

**STEP 3:**

A^{2} = 6^{2} = 36

There is a carry of โ8โ from the previous step

Therefore, [36+8] = 44

The final answer is **67 ^{2} = 4 4 8 9**

To find the square of a number which is a** multiple of โ5โ**

**AB ^{2} = [Aร next number] B^{2}**

25^{2} = [2ร3] 5^{2}

= [6] 25 = 625

i.e., AB^{2} where **B=5**

**AB ^{2} = [Aร next number] B^{2}**

For example, **85 ^{2}** = [8ร9] 25 =

**7225**

**ย ย 115**^{2ย }= [11ร12] 25 = **13225**

**155 ^{2}** = [15ร16] 25 =

**24025**

**This method can be followed for all numbers divisible by 5**

**Easy Method to Take Square Root for a Number**

**Numerical Ability – SQUARE ROOT**

**POINTS TO REMEMBER:**

- When 22 = 4, then โ4 = 2
- Here 4 is the
**square**of 2 - 2 is the
**square root**of 4 - A
**Square**of a number can**never**end with**2, 3, 7 and 8**

**Table 1:**

Oneโs digit of a square |
Oneโs digit of the square root |

1 | 1 or 9 |

4 | 2 or 8 |

5 | 5 |

6 | 4 or 6 |

9 | 7 or 3 |

To find the square of a number which is a** multiple of โ5โ**

25^{2} = [2ร3] 5^{2}

= [6] 25 = 625

i.e., AB^{2} where **B=5**

**AB ^{2} = [Aร next number] B^{2}**

For example,** 85 ^{2}** = [8ร9] 25 =

**7225**

**115**= [11ร12] 25 =

^{2}**13225**

**155**= [15ร16] 25 =

^{2}**24025**

**This method can be followed for all numbers divisible by 5**

**TYPE 1:**

To find the square root of a 3-digit number

EXAMPLE: **โ841**

**STEP 1:** Consider the oneโs digit of the given number i.e., 1

From Table 1, if the oneโsย digit of the square is โ1โ then the square root would either end with **โ1โ or โ9โ**

**STEP 2:** Always** ignore the tenโs digit** of the given number

**STEP 3:** Now the remaining number other than the oneโs and the tenโs digit in the given number is **โ8โ.ย **Consider a **square-root of a square** which is **nearer to** as well as **lesser than** โ8โ.

Here it is โ4โ which is nearer to as well as lesser than โ8โ. Hence the square root of 4 i.e., **โ2โ** is taken.

**STEP 4:** we already know the oneโs digit of the square root to be either 1 or 9 from STEP 1

Therefore the square root of โ841โ lies between **21** and **29**

**STEP 5:**

Take a number **divisible by โ5โ** between 21 and 29, that is **โ25โ**

25^{2} = [2ร3] 25 = **625**

Now **625 < 841**

252 is itself lesser than 841. Then **21**^{2} will be **much lesser** than **841.**

Therefore, the remaining option is โ29โ

**โ841 = 29**

^{2}

**TYPE 2:**

To find the square root of a 4-digit number

EXAMPLE:** โ8464**

**STEP 1:** Consider the oneโs digit of the given number i.e., 4

From Table 1, if the oneโs digit of the square is โ4โ then the square root would either end with** โ2โ or โ8โ**

**STEP 2:** Always **ignore the tenโs digit** of the given number

**STEP 3:** Now the remaining numbers other than the oneโs and the tenโs digit in the given number is **โ84โ**

Consider a **square-root of a square** which is **nearer to** as well as **lesser than** โ84โ.

Here it is **โ81โ** which is nearer to as well as lesser than โ84โ. Hence the square root of 81 i.e., **โ9โ** is taken.

**STEP 4:** we already know the oneโs digit of the square root to be either 2 or 8 from STEP 1

Therefore the square root of โ8464โ lies between** 92** and **98**

**STEP 5:**

Take a number **divisible by โ5โ** between 92 and 98, that is โ95โ

952 = [9ร10] 25 = **9025**

Now **9025 > 8464**

952 is itself greater than 8464. Then **98 ^{2}** will be

**much greater**than

**8464**

Therefore, the remaining option is โ92โ

**โ8464 = 92**

**TYPE 3:**

To find the square root of a 5-digit number

EXAMPLE: **โ18769**

**STEP 1:** Consider the oneโs digit of the given number i.e., 9

From Table 1, if the oneโs digit of the square is โ9โ then the square root would either end with** โ3โ or โ7โ**

**STEP 2:** Always **ignore the tenโs** **digit** of the given number

**STEP 3:** Now the remaining numbers other than the oneโs and the tenโs digit in the given number is โ187โ

Consider a **square-root of a square** which is **nearer to** as well as** lesser than** โ187โ

Here it is** โ169โ** which is nearer to as well as lesser than 187. Hence the square root of 169 i.e., **โ13โ** is taken

**STEP 4:** we already know the oneโs digit of the square root to be either 3 or 7 from STEP 1

Therefore the square root of โ18769โ lies between **133** and **137**

**STEP 5:**

Take a number **divisible by โ5โ** between 133 and 137, that is **โ135โ**

135^{2} = [13ร14] 25 = **18225**

Now **18225 < 18769**

1352 is itself smaller than 18769. Then **133 ^{2}** will be much lesser than

**18769.ย**Therefore, the remaining option is โ137โ

**โ18769 = 137**

**Shortcut for the Multiplication of 2 Digit & 3 Digit Numbers –ย Numerical Ability**

**MULTIPLICATION OF 2-DIGIT NUMBERS (10-99):**

**EXAMPLE 1:**

78ร65 =?

**GIVEN:**

N1 = 78; N2 = 65

**SOLUTION:**

**STEP 1:**Oneโs digit of the product is obtained by multiplying the oneโs digits of N1 and N2

**Oneโs digit of product = [Oneโs digit of N1 ร Oneโs digit of N2]**

= [8ร5]

=**4**0

โ4โ is taken as carry to the Step 2 i.e.,**C1 = 4**

**STEP 2:**

**Tenโs digit of product = [Tenโs digit of N1รOneโs digit of N2] +ย ****[Oneโs digit of N1รTenโs digit of N2] + C1**

= [7ร5] + [6ร8] + 4

= 35 + 48 + 4

=** 8**7

โ8โ is taken as the carry to Step 3 i.e.,** C2 = 8**

Add the Products of the above 2-steps. At the end of Step 2 we have

**STEP 3:**

To calculate the **hundredth and the thousandth digit** **of the product**

**= [Tenโs digit of N1 ร Tenโs digit of N2] + C2**

= [7ร6] + 8

= [42] + 8

= 50

Therefore, 78ร65 = 5070.

**Shortcut for the Multiplication of 2 Digit & 3 Digit Numbers –ย Numerical Ability**

**MULTIPLICATION OF NUMBERS BETWEEN (90-99):**

The product can be determined by a **simple multiplication** and a **subtraction.**

**EXAMPLE:**

98 ร 97 = ?

Here, these numbers are nearer to **โ100โ** in the number system.

So their difference from โ100โ is considered, i.e., [**D1**=100-98=02] & [**D2**=100-97=03]

**GIVEN:**

**No.1** = 98;** No.2** = 97

Difference,** D1** = 02; **D2** = 03

**SOLUTION:**

The multiplication is done between 2-digit numbers so the **product** will definitely be a **4-digit number.**

The **oneโs** and** tenโs** digit of the **product** will be the **product of D1 and D2**

i.e., D1รD2 = 02ร03 = 06

Similarly, the **hundredth** and **thousand** digit of the **product** will be the** difference between No.2 andย D1. i.e.,** [No.2-D1 = 97-02=95]

Therefore, **98ร97 = 9506**

This method can be applied t**o multiply 2-digit numbers whose difference from โ100โ can be multipliedย with each other easily.**

88 ร 89 = ?

Here, No.1=88; No.2=89; D1=100-88=12; D2=100-89=11

Here, 12 ร 11 = **1**32, we know that the product is a 4-digit number so** โ1โ is taken as carry**

For the hundredth and thousand digits of the product,** No.2 โ D1 = 89-12=77**

Now we have one carry from the previous step โ1โ, that is added to this difference [77+**1**=78]

Therefore, **88ร89=7832**

**MULTIPLICATION OF NUMBERS BETWEEN (101-110):**

Here the product can be determined by a **simple multiplication** and an **addition.**

**EXAMPLE:**

108 ร 107 = ?

Here, these numbers are nearer to **โ100โ** in the number system.

So their difference from โ100โ is considered, i.e., [**D1**=108-100=08] &

[**D2**=107-100=07]

**GIVEN:**

**No.1** = 108; **No.2** = 107

Difference, **D1** = 08;** D2** = 07

SOLUTION:

The multiplication is done between 3-digit numbers below 200(<200) so the **product** will definitely be a **5-digit number.**

Here, the method is same as that of 2-digit multiplication except for a small change.

**STEP 1: Product of D1 and D2 is same i.e., **[D1รD2=08ร07=56]

**STEP 2:** Since the numbers are** greater than 100,** the **sum of No.2 and D1** is taken

**[No.2+D1=107+08=115]**

Therefore, **108ร107=11556**

Similarly, if D1 and D2 are **more than 10,** then the **carry over** from the product of **D1รD2** is added to the sum of **No.2+D1**

For example, **111ร112=?**

Here D1=11; D2=12 **(>10)**

Therefore,** D1รD2=11ร12=1**32

In step 2, No.2+D1=112+11=123+carry=123+**1**=124

So, **111ร112=12432**

**Numerical Ability – Shortcut for the Multiplication of 3 Digit & 4 Digit Numbers**

**MULTIPLICATION OF NUMBERS BETWEEN (900-999):**

The product can be determined by a** simple multiplication** and a **subtraction.**

**EXAMPLE:**

**998 ร 997 **= ?

Here, these numbers are nearer to **โ1000โ** in the number system.

So their difference from โ1000โ is considered, i.e., [**D1**=1000-998=002] & [**D2**=1000-997=003]

**GIVEN:**

**No.1** = 998; **No.2** = 997

Difference, **D1** = 002; **D2** = 003

**SOLUTION:**

The multiplication is done between 3-digit numbers so the **product** will definitely be a **6-digitย number.**

The oneโs**, tenโs** and **hundredth** digit of the **product** will be the **product of D1 and D2**

i.e., D1รD2 = 002ร003 = 006

**Similarly, the 1st 3 digits of the product will be the difference between No.2 and D1. i.e., [No.2-D1 = 97-02=95]**

Therefore, **998ร997 = 995006**

This method can be applied **to multiply 3-digit numbers whose difference from โ1000โ can beย multiplied with each other easily.**

988 ร 989 = ?

Here, No.1=988; No.2=989; D1=1000-988=12; D2=1000-989=11

Here, 12 ร 11 = 132

For the 1st 3 digits of the product, **No.2 โ D1 = 989-12= 977**

Therefore, **988ร989=977132**

**MULTIPLICATION OF NUMBERS BETWEEN (1001-1010):**

Here the product can be determined by a **simple multiplication** and an **addition.**

**EXAMPLE:**

1008 ร 1007 = ?

Here, these numbers are nearer to **โ1000โ** in the number system.

So their difference from โ1000โ is considered, i.e., [**D1**=1008-1000=008] & [**D2**=1007-1000=007]

**GIVEN:**

**No.1** = 1008;** No.2** = 1007

Difference,** D1** = 008; **D2** = 007

**SOLUTION:**

The multiplication is done between 4-digit numbers so the **product** will definitely be a** 7-digit number.ย ย **

Here, the method is same as that of 3-digit multiplication except for a small change.

**STEP 1:** product of D1 and D2 is same i.e.,** [D1รD2=008ร007=056]**

**STEP 2:** Since the numbers **are greater than 1000,** the **sum of No.2 and D1** is taken

**[No.2+D1=1007+08=1015]**

Therefore,** 1008ร1007=1015056**

**Fast and Easy Method to Take Cube Root –ย Numerical Ability**

**POINTS TO REMEMBER:**

- When 2
^{3}= 8, then (8)^{1/3}= 2 - Here 8 is the
**cube**of 2 - 2 is the
**cube root**of 8

**Table 1:**

Oneโs digit of a cubeย |
Oneโs digit of the cube root |

1 | 1 |

2 | 8 |

3 | 7 |

4 | 4 |

5 | 5 |

6 | 6 |

7 | 3 |

8 | 2 |

9 | 9 |

0 | 0 |

**TYPE 1:** To find the cube root of a 4-digit Number

**EXAMPLE:** (9261)^{1/3}

**STEP 1:** Consider the oneโs digit of the given number i.e., 1

From Table 1, if the oneโs digit of the cube is โ1โ then the cube root would also end with โ1โ

**STEP 2:** Always** ignore the tenโs** and the **hundredth digits** of the given number i.e., ignore โ2โ and โ6โ in the given number

**STEP 3:** Now the remaining number other than the oneโs, tenโs and the hundredth digit in the givenย number is โ9โ

Consider a **cube-root of a cube** which is nearer to as well as lesser than โ9โ.

Here it is โ8โ which is nearer to as well as lesser than โ9โ. Hence the cube root of 8 i.e., โ2โ is taken

**STEP 4:** we already know the oneโs digit of the cube root is **1** from STEP 1

Therefore the cube root of โ9261โ is 2**1**

Therefore **(9261) ^{1/3} = 21**

**TYPE 2:**

To find the cube root of a 5-digit number

**EXAMPLE:** (32768)^{1/3}

**STEP 1:** Consider the oneโs digit of the given number i.e., 8

From Table 1, if the oneโs digit of the cube is โ8โ then the cube root would end with โ2โ

**STEP 2:** Always ignore the tenโs and the hundredth digits of the given number i.e., โ6โ and โ7โ here

**STEP 3:** Now the remaining numbers other than the oneโs, tenโs and the hundredth digits in the given number is โ32โ

Consider a **cube-root of a cube** which is nearer to as well as lesser than โ32โ.

Here it is โ27โ which is nearer to as well as lesser than โ32โ. Hence the cube root of 27 i.e., โ3โ is taken

**STEP 4:** we already know the oneโs digit of the cube root to be **2** from STEP 1

Therefore the cube root of โ32768โ is 3**2**

Therefore **(32768) ^{1/3} = 32**

**Shortcut Numerical Ability Rules for Basic Multiplication**

**Type-1:** If the unit figure is same and the sum of the tens figure is 10, then follow the below method.

**General Shortcut Method:**

[Tens fig. ร Tens fig. + Unit fig.] [Unit fig ร Unit fig]

**Example:**

86 ร 26 = [8 ร 2 + 6] [6 ร 6] = [22] [36]; so answer is: 2236.

**Note:** Here, the unit figure denotes the number that present in the ones digit (6, 6), and tens figure denotes the number that present in the tens digit (8, 2).

**Type-2:** If the sum of the unit figure is 5 and the tens figure are equal. Then follow the below method.

**General Shortcut Method:**

[(Tens figure)2 + ยฝ ร Tens figure] [Unit fig. ร Unit fig]

**Example:**

83 ร 82 = [82 + ยฝ ร 8] [ 3 ร 2 ] = [68] [06] So the answer is: 6806

**Type-3:** If the unit figures are same and the sum of tens figures is 5.

**General Shortcut Method:**

[Tens fig ร Tens fig + ยฝ ร Unit fig] [(Unit fig.)2]

**Example:**

36 ร 26 = [3 ร 2 + ยฝ ร 6] [62]

= [9] [36]; Answer is 936.

**Type-4:** If the unit figures are 5 and difference between the tens figures is 1 then the rule is,

**General Shortcut Method:ย ย **[(Larger tens fig + 1) ร (Smaller tens fig)] [75]

**Example:**

35 ร 45 = [(4 + 1) ร 3] [75]

= [15] [75]; So the Answer is, 1575.

**MULTIPLICATION BY 11 WITH ANY NUMBER AND ANY DIGITS –ย Numerical Ability**

**TWO DIGIT NUMBERS:**

**TYPE I:** When the sum of the oneโs and tenโs digit of the number is less than 10 (i.e. 0 to 9)

**For example,** 43ร11 where the sum of the digits is less than 10, [4+3=7] this method can be used.

Now **43ร11**

**ย ย ย ย 4 _ 3**

The oneโs and the tenโs digit of the number will be the last and the first digit of the product

respectively.

**43ร11**

**ย 4 7 3**

The tenโs digit of the product will be sum of the digits of the number, here **[4+3=7]**

Therefore, 43 ร 11 = 473.

Similarly, 36 ร 11 = 396, where** [3+6=9]**

53 ร 11 = 583, where **[5+3=8]**

**TYPE II:** When the sum of the oneโs and tenโs digit of the number is either 10 or more than 10 (i.e. 10 or 10<)

**For example,** 28 ร11 where the sum of the digits is 10, [2+8=10] this method can be used.

Now **28ร11**

**ย ย ย ย ย _ _ 8**

The oneโs digit of the number will be the oneโs digit of the product.

**28ร11**

**_ 0 8**

The tenโs digit of the product will be sum of the digits of the number, here **[2 + 8 = 1 0]โ**

**ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย Carryover**

The hundredth digit of the product will be the sum of tenโs digit of the number and the carry over.

**[2 + 1 = 3]**

Therefore, **28ร11**

**ย ย ย ย ย ย ย ย ย 3 0 8**

Similarly, 85 ร 11 = 935ย โ {9[8+1] 3[8+5=13] 5}

99 ร 11 = 1089 โ {10[9+1] 8[9+9=18] 9}

**THREE DIGIT NUMBERS:**

**TYPE I** (Sum of digits < 10):

**For example:**

352 ร 11

Here the sum of the digits, 3+5=8ย &ย 5+2=7โ

**ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย Both the sums are less than 10**

Now, **352 ร 11**

**3 _ _ 2**

Similar to 2-digit multiplication the first and the last digit of the number will be the first and the last

digit of the product.

The tenโs digit of the product = [Oneโs digit of the no. + Tenโs digit of the no.]

= [5 + 2 = 7]

**352 ร 11**

**ย 3 _ 7 2**

The hundredth digit of the product = [Hundredth digit of the no. + Tenโs digit of the no.]

= [3 + 5 = **8**]

**352 ร 11**

**ย 3 8 7 2**

Therefore, 352ร11 = 3872.

Similarly, 236 ร 11 = 2596 โ {2 5[2+3=5] 9[3+6] 6}

123 ร 11 = 1353 โ {1 3[1+2=3] 5[2+3] 3}

**TYPE II (Sum of digits >10):**

**For example:**

756 ร 11

Here the sum of the digits, 7+5=12 &ย 5+6=11โ

**ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย Both the sums are more than 10**

Now, **756 ร 11**

**ย ย ย ย ย _ _ _ 6**

The oneโs digit of the product is same as the oneโs digit of the number.

Tenโs digit of the product = [Oneโs digit of no. + Tenโs digit of no.]

= [5 + 6 = 1 1]

**Carry over 1**

**756 ร 11**

**ย _ _ 1 6**

Hundredth digit of the product = [Tenโs digit of no.+ Hundredth digit of no.] + Carry over 1

= [7 + 5] + 1 = 1 3

**Carry over 2**

**756 ร 11**

**ย _ 3 1 6**

Thousand digit of the product = [Hundredth digit of the number + Carry over 2]

= [7 + 1 = 8]

**756 ร 11**

**ย 8 3 1 6**

Therefore, 756 ร 11 = 8316

Similarly, 999 ร 11 = 10989 โ {10[9+1] 9[9+9+1=19] 8[9+9=18] 9}

**Shortcut Numerical Ability Rules for Basic Subtraction**

**Rule-I: Borrowing and Paying Back Method:**

This method is the quickest method of subtraction. This method is also called equal additions method.

**Example (1):** Suppose we have to subtract 55 from 91. Mentally we have to increase the number to be subtracted to the nearest multiple of 10 i.e., increase 55 to 60 by adding 5 to it. Mentally increase the other quantity by the same amount i.e., by 5. Therefore, the problem is 96 minus 60 i.e., our answer is 96 โ 60= 36.

**Example (2):** Sometimes it is useful to increase the number to be subtracted to the nearest multiple of 100.

for example 442 โ 179. Therefore 179 becomes 200 by adding 21 and 442 becomes 463 by adding 21.ย Then the problem becomes 463 โ 200= 263. Now we see that 463 โ 200 is easier than 442 โ 179. The result is same as 263.

**Example (3):** Another example is 2326 โ 1875. Here 1875 becomes 2000 by adding 125 and 2326 becomes 2451 by adding 125. The number becomes 2451 โ 2000= 451. Here the subtraction 2451 โ 2000 is easier than the subtraction 2326 โ 1875. The answer of both is same 451.

**Example (4):** The subtraction of 3786 โ 2998. Here 2998 becomes 3000 by adding 2 and 3786 becomes 3788 by adding 2. The problem of 3788 โ 3000 is easier than 3786 โ 2998 and our answer is 788. This answer is same for both the problems.

**Rule: II. Double Column Addition and Subtraction Method:**

This following method works when there is a series of additions and subtractions are to be performed in a line.

**Example (1):**

1026

– 4572

+5263

– 2763

+8294

_____________

**Explanation:** We have to look the signs given before the numbers and then start adding and adding and subtracting from the top right position.

**Step I:** First Double Column

26 โ 72= – 46, – 46 + 63= 17,

17 โ 63= – 46, – 46 + 94= 48.

Answer is 48ย โ Step-I

**Step II:** Second Double Column:

10 โ 45= – 35, – 35 + 52= 17,

17 โ 27= – 10, – 10 + 82= 72.

Answer is 72ย โ Step-II

Now Combine the Step II and Step I.

Answer will be 7248.

You will get the same answer if you also use the normal method.

**Example (2):**

7676

– 1431

+5276

– 3489

+1546

_____________

**Explanation:** We have to look the signs given before the numbers and then start adding and adding and subtracting from the top right position.

**Step I: First Double Column**

76 โ 31= 45, 45 + 76= 121 (here in the 121 take the last two digits from 121 i.e., 21)

21 – 89= – 68, – 68 + 46= -22 (Here the answer comes in minus so add 100 with the answer)

100 + (- 22) = 78.

Answer is 78ย โ Step I

**Step II: Second Double Column**

76 – 14= 62, 62 + 52= 114 (Take the last two digits from 114 i.e., 14)

14 โ 34= – 20, – 20 + 15= – 5 (Here the answer comes in minus so add 100 with the answer)

100 + (-5) = 95

Answer is 95ย โ Step II.

Now Combine the Step II and Step I.

The answer will be 9578.

**Shortcutย Numerical Ability Rules for Basic Division**

**1.) DIVISIBLE BY 2:**

A number will be divisible by 2, if the unit digit in the number is 0, 2, 4, 6 and 8.

**Example:** Numbers like, 56456, 32658, 89846 are divisible by 2.

**2.) DIVISIBLE BY 4:**

A number will be divisible by 4, if the last two digits of the number is divisible by 4.

**Example:** Numbers like 56536 is divisible by 4, because the last two digits of this number is divisible by 4 and the number 546642 is not divisible by 4 because the last two digits of this number is not divisible by 4.

**3.) DIVISIBLE BY 6:**

A number will be divisible by 6, if that number is divisible by both 2 and 3.

**Example:** 36 is divisible by 6 because 36 is divisible by both 2 and 3.

**4.) DIVISIBLE BY 8:ย **A number will be divisible by 8, if the last three digits of that number are divisible by 8.

**Example:** 565144 is divisible by 8 because the last three digits 144 is divisible by 8. And the number 554314 is not divisible by 8 because the last three digits 314 is not divisible by 8.

**5.) DIVISIBLE BY 5:**

A number will be divisible by 5 if the unit digit is either 0 or 5.

**Example**: Numbers like 565520 and 898935 are divisible by 5.

**6.) DIVISIBLE BY 3:**

A number will be divisible by 3, if the sum of the digits in the number is divisible by 3.

**Example:** 658452 is divisible by 3 because the sum of the numbers is divisible by 3, 6+5+8+4+5+2= 30, which is divisible by 3.

The number 456455 is not divisible by 3, because the sum of the number is not divisible by 3. 4+5+6+4+5+5= 29 this is not divisible by 29.

**7.) DIVISIBLE BY 9:**

A number will be divisible by 9, if the sum of the digits in the number is divisible by 9.

**Example:** 898686 is divisible by 9 because the sum of the numbers is divisible by 9,

8+9+8+6+8+6= 45, this is divisible by 9.

**8.) DIVISIBLE BY 11:**

A number will be divisible by 11, if the difference of the sum of the digits in the Odd places and Sum of the digits in the Even places, is either zero or divisible by 11.

**Example:** 502678 is divisible by 11 because, the sum of the digits of the odd places, 5+2+7= 14, sum of the digits in the even places, 0+6+8=14, the difference is 14-14=0, so this number is divisible by 11.

**9.) DIVISIBLE BY 12:**

A number is divisible by 12, if the number is divisible by both 3 and 4.

Example: 144 is divisible by 12, because it is divisible by both 3 and 4.

**10.) DIVISIBLE BY 10:
**Any number that ends with zero will be divisible by 10.

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