Numerical Ability Test - 1

Concept Used:         

Follow the BODMAS rule according to the table given below:

Calculation:

2 × 8 + (7 × 4) ÷ 2 – 24

⇒ 2 × 8 + 28 ÷ 2 – 24

⇒ 2 × 8 + 14 – 24

⇒ 16 + 14 – 24

⇒ 30 – 24

⇒ 6

∴ The required value is 6

Given:

3-x × 9x - 3 = 81

Formula Used:

am × an = am + n

Calculation:

Considering the given statement

3-x × 9x - 3 = 81

⇒ 3-x × 3²⁽x - 3) = 3⁴

⇒ 3^{-x + 2x - 6} = 3⁴

By comparing the exponents

- x + 2x - 6 = 4

⇒ x = 4 + 6

⇒ x = 10

Formula Used:

Dividend = Divisor × Quotient + Remainder

Calculation:     

Let the quotient be Q

Number = 731 Q + 37 = 17 × 43Q + 34 + 3 = 17(43Q + 2) + 3

When this number is divided by 17 then, the remainder is 3

Remainder = 3 

Given:

The number = 827, 1149 and 1310

Concept used:

HCF - The largest number or quantity that is a factor of each member of a group of numbers or quantities 

Calculation:

Take the relative difference of the given number

HCF of 827, 1149 and 1310

⇒ (1149 – 827) = 322

⇒ (1310 – 1149) = 161

⇒ (1310 – 827) = 483

Now,

HCF of 322, 161 and 483 is

⇒ 322 = 2 × 161

⇒ 161 = 1 × 161

⇒ 483 = 3 × 161

The HCF is 161

∴  The largest number that divides 827, 1149 and 1310 to leave the same remainder in each case is 161

Concept used:

HCF or the Highest Common Factor of two or more numbers is the greatest factor that divides the numbers.

Calculation:

(x2 + 4x + 4) 

⇒ (x2 + 2 × x × 2 + 2²)

⇒ (x + 2)² 

And, (3x2 - 12)

⇒ 3(x² - 4)

⇒ 3(x² - 2²)

⇒ 3(x + 2)(x - 2)

∴ The HCF of (x2 + 4x + 4) and (3x2 - 12) is (x + 2)

Given:

Four bells ring in a temple at intervals of 12, 16, 24, and 36 minutes respectively.

They are started playing at a continuous interval from 6 o'clock

Concept used:

The LCM of 12, 16, 24, and 36 minutes will give us the next time they will all ring together.

Calculation:

LCM (12, 16, 24, 36) = 144

So, the bells will play together after 144 mins i.e. 2 hours 24 mins.

So, the bells will play together again at = (6.00 + 2.24) = 8.24 AM

∴ At 08.24 AM they will play together again.

Concept:

LCM of fractions = (LCM of Numerators)/(HCF of Denominators)

Calculation:

LCM of Numerators = LCM (3, 8, 6) = 24

HCF of Denominators = HCF (7, 5, 7) = 1

∴ LCM of \(\left( \frac {3}{7}, \frac{8}{5}, \frac{6}{7} \right)\) 

⇒ 24/1 = 24

∴ The required answer will be 24.

Additional Information 
\(HCF\ of\ the\ fractions\ =\ {The\ HCF\ of\ the\ Numerators \over The\ LCM\ of\ the\ denominators}\)

Given:

6, 7, 8, 9, and 12.

Concept used:

The least common multiple (LCM) of a set of numbers is the smallest number that leaves 0 as the remainder each time when divided by any of those numbers.

Calculation:

LCM (6, 7, 8, 9 12) = 504

So, 504 is the smallest number that leaves 0 as the remainder each time when divided by any of 6, 7, 8, 9, and 12.

So, the required number = (504 + 2) = 506

∴ 506 will be the smallest number which when divided by 6, 7, 8, 9, and 12 leaves the remainder 2 each time.

Calculation:

8² = 64

(0.08)² = 0.0064

∴ √0.0064 = 0.08

15² = 225

∴ √2.25 = 1.5

17² = 289

∴ √2.89 = 1.7

⇒ 0.08 + 1.5 + 1.7 =  3.28

Given:

The ratio of the two numbers = 2 : 3

LCM of the two numbers = 48

Calculation:

Let the two numbers be 2y and 3y. 

LCM (2y, 3y) = 6y 

⇒ 6y = 48 

⇒ y = 8 

Now, The sum of numbers = (2y + 3y)

⇒ 5y 

⇒ 5y = 5 × 8

⇒ 40

∴ The sum of the number is 40.

∴ The correct answer is None of the above