Mettl Number System Quiz 1

Let the number be z. then, z – 36 = 86 – z

⇒  2z = 86 + 36 ⇒ 2z = 122.

⇒  z = 61.

Hence, the required number is 61.

Let, the number be z, Then, 7z – 15 = 2z + 10

⇒  5z = 25  ⇔ z = 5.

Hence, the required number is 5.

We can easily solve this question by applying the 'Hit & Trial' approach.

The condition says that sum of the numbers and its reciprocal must be (13/6) and in

the 1st option, the sum of

\frac{5}{3}(\text{ the num }) + \frac{3}{5}(\text{ its reciprocal }) will give us \frac{34}{15}

Option 'A'  hence gets eliminated.

the 2nd option, the sum of

2 + \frac{1}{2} = \frac{5}{2}

Option 'B' hence gets eliminated too.

the 3rd option, the sum of

\frac{3}{2} + \frac{2}{3} = \frac{13}{6}     .

Option 'C' hence is the correct answer.

Let the number be z and (184 – z). then,

\Rightarrow \frac{z}{3} - \frac{(184 - z)}{7} = 8.

⇒  7z – 3(184 – z) = 168

⇒  10z = 720

⇒   z = 72.

Let the numbers be x and y. then,

x – y = 11       .....(1)  and,

\Rightarrow \frac{1}{5}(x + y) = 9

⇒  x + y = 45    ...(2)

Adding (1) and (2) we get :   2x = 56 or x = 28. putting x = 28 in (1), we get : y = 17.

Hence, the numbers are 28 and 17.

Let the number be x and y. then, x + y = 42 and xy = 437.

x - y = \sqrt{(x + y)^2 - 4xy}

\Rightarrow = \sqrt{(42)^2 - 4\times 437} \Rightarrow \sqrt{1764 - 1748}

\Rightarrow = \sqrt{16} = 4

So, the required difference  = 4.

Let the numbers be x and (15 – x).
Then, x² + (15 – x)² = 113 ⇔ x² + 225 + x² – 30x = 113
⇒ 2x² – 30x + 112 = 0 ⇔ x² – 15x + 56 = 0
⇒ (x – 7)(x – 8) = 0 ⇔ x = 7 and x = 8.
So, the numbers are 7 and 8

Let the four consecutive even numbers be x, x + 2, x + 4 and x + 6.

Then, sum of these numbers =  (27 × 4)  = 108.

So, x + (x + 2) + (x + 4) + (x + 6) = 108  or  4x  =  96   or  x = 24.

So, the largest number  = (x + 6) = 30.

Let the number be x , x + 2 and x + 4.

Then, x² + (x + 2)² + (x + 4)² = 2531 ⇔  3x² + 12x – 2511 = 0

⇒ x²  + 4x – 837 = 0 ⇔   (x – 27)(x + 31) = 0  ⇔  x = 27.

Hence, the required numbers are 27, 29 and 31.

Let the number be x and y, such that x > y .

Then, 3x – 4y = 5   ....(1)  and  (x + y) – 6 (x – y) = 6

⇒  – 5x + 7y  = 6 .....(2)

Solving (1) and (2), we get : x = 59 and y = 43.

Hence, the required numbers are 59 and 43.