Practice Questions

√(25% of 64 × 16) + 40 

= √(25/100 × 64 × 16) + 40

= √(1/4 × 64 × 16) + 40

= 16 + 40

= 56

5.51 ÷ 0.0019

= {551 ÷ 100}/{19 ÷ 1000}


= 551 × 100/19

= 2900

24 ÷ 4 ×(3 +3) ÷2 

= 24 ÷ 4 × 6   ÷2 

= 6  × 6   ÷2 

= 18

25% of 890 + 29% of 383 - 22% of 1422 = ?

25(890)/100 + 29(383)/100 - 22(1422)/100 = ?

222.5 + 111.07 - 312.84 = ?

20.73 = ?

18 ÷ 9 × 4 + 36 of 6 ÷ 1 – 42

⟹ 2 × 4 + 216 – 42

⟹ 224 – 42 = 182

75 of 3 + (49 ÷ 7) × 6 + 105 ÷ 15 of 7

⟹ 225 + 7 × 6 + 1

⟹ 225 + 42 + 1 = 268

768 ÷ 8 of 6 × [126 ÷ 9 × (19 – 5) of 26 – (9-4)] ÷ 3

⟹ 768 ÷ 48 × [14 × 364 – 5] ⟹ 16 × 5091 ÷ 3 = 16 × 1697 = 27152

(320 + 342 + 530 + 915) ÷ (20 + 22 – x + 18) = 43

⟹ 2107= 43 × (60-x)

⟹ x = 60 – 49

⟹ x = 11

2.5 × 2 + 3 ÷ 3 + 6 

= 5 + 1 + 6

= 12

25-[7+{27-(18-6)}]

= 25-[7+{27-12}]

= 25-[7+15]

= 25-22

= 3

(⅖) × 5 ¼

Here, 5 ¼ is a mixed fraction.

Let us convert this mixed fraction into an improper fraction.

5 ¼ = [(5 × 4) + 1]/4 = 21/4

Thus, (⅖) × 5 ¼ = (⅖) × (21/4) = 21/10

5/129

Here, numerator = 5

Denominator = 129

Reciprocal of 5/129 = 129/5

The product of 5/129 and its reciprocal = (5/129) × (129/5) = 1.

2 x ? - 6 = 676/26

 2 x ? - 6 = 26

2 x ? = 26 + 6 = 32

? = 32/2 = 16

4/5 ÷ 2/3

= 4/5  × 3/2

= 12/10

= 6/5

Convert each fraction to a decimal to easily compare:

2/3 = 0.666...

3/4 = 0.75

5/6 = 0.833...

7/8 = 0.875

As we know a number with unit digit 2 have repeating cycle of 2,4,8,6 after every fourth power as power is 173 or (172+1) where till 172 , 43rd cycle will get complete and next unit digit will be 2

Unit digit of a product = Unit digit of the product of unit digits of individual number

A number is divisible by 4 if its last two digits are divisible by 4, or if the last two digits are 00. A number is divisible by 8 if its last three digits are divisible by 8, or if the last three digits are 000

A number is divisible by 4 if its last two digits are 00 or a multiple of 4

So here XX = 60

A number is divisible by 9,if the sum of all digit is divisible by 9.

A number is divisible by 3,if the sum of all digit is divisible by 3.

So here we have to consider the number X in which the sum is 9 

 72354X2 = 7 + 2 + 3+ 5 + 4 +X + 2

= 23 + X

So X = 4

A number is divisible by 15 if the number is divisible by both 3 and 5

A number is divisible by 9 
If the sum of the digits of given number is divisible by 9 

So here

936819 = 9 + 3 + 6 + 8 + 1 + 9

= 36

= multiple of 9

If a number is divisible by 9, the sum of digits is also divisible by 9

9 + 8 + 5 + x + 8 + 6 + 5 = 9x

41 + x = 9x

Nearest value of 9x must be 45

41 + x = 45

x = 4

A number is divisible by 4 if the last 2 digits is divisible by 4

As we know, a number is divisible by 4, if the last two digits of the given number are exactly divisible by 4.

In the given number 2024, the last 2 digits are 24.

Here, 24 is completely divisible by 4.

i.e., 24/4 = 6.

Therefore, 2024 is divisible by 4.

For a number to divisible by 4, last two digits of the number should be divisible by 4. But 71 is not divisible by 4. Hence the number is not divisible by 4.
For a number to be divisible by 3, the sum of the digits should be divisible by 3. But, 7+3+8+6+0+7+1=32 is not divisible by 3. Hence the number is not divisible by 3.
For a number to be divisible by 9, the sum of the digits should be divisible by 9. But, 7+3+8+6+0+7+1=32 is not divisible by 9. Hence the number is not divisible by 9.
For a number to be divisible by 11, the difference between the sum of digits in odd places and the sum of digits in even places should be either 0 or a number divisible by 11. (7+8+0+1) – (3+6+7) = 0, which is divisible by 11. Hence the number is divisible by 11.

We know that, for a number to be divisible by 11, the difference between the sum of digits in odd places and the sum of digits in even places should be either 0 or a number divisible by 11.
(2+X+5) – (3+Y+9) = 0 or multiple of 11
X – Y – 5 = 0 or multiple of 11
If X=6 and Y=1, X – Y – 5 = 0 and hence divisible by 11.
If X=1 and Y=7, X – Y – 5 = -11, which is a multiple of 11 and hence divisible by 11.
XY=17 cannot be the answer, because 23XY70 is a number with all distinct digits and 7 is already present in the number.
Therefore, XY is 61.

We know that, for a number to be divisible by 11, the difference between the sum of digits in odd places and the sum of digits in even places should be either 0 or a number divisible by 11.

(1+4+X+2) – (2+5+4) = 0 or multiple of 11

X – 4 = 0

Therefore, X = 4.

A number is divisible by 15 if it is divisible by both 3 and 5

If the number is divisible by both 2 and 7 then it is divisible by 14

If the number is divisible by both 3 and 4 then it it is divisible by 12

A number is divisible by 6 if it is divisble by both 2 and 3

A number is divisible by 2 if it's last digit is an even number

Co-prime numbers, also known as relatively prime numbers, are two numbers that have only 1 as their greatest common divisor (GCD) or highest common factor (HCF)

A number is divisible by 3 if the sum of the digits is a multiple of 3

A number is divisible by 5 if it's last digit is either 5 or 0

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