Math, asked by adlakhamadhur2, 10 months ago

3,1,8 2,6,2 4,7,5 65 ? 54
a 58 b 60 c 68 d 62

Answers

Answered by poojan
15

Given data:

3      1      8

2      6     2

4      7     5

65    ?    54

Options:

(1) 60

(2) 58

(3) 68

(4) 62​

To find:

The missing number.

Solution:

By option verification.

In each column, the sum of singled digits of each row goes in the series 20, 22, 24,...

We need to find the option that satisfies the total to get 22.

Column 1:  3 + 2 + 4 + (6 + 5) = 20

Column 2: 1 + 6 + 7 + (x + y) = 22

Column 3: 8 + 2 + 5 + (5+4) = 24

From column 2,

1 + 6 + 7 + (x + y) = 22

14 + (x + y) = 22

(x + y) = 22 - 14

(x + y) = 8

By option verification, find which value on adding their individual digits gives 8 as a sum.

Option 1: 60 = 6+0 = 6 which is not equal to 8 (incorrect)

Option 2: 58 = 5+8 = 13 which is not equal to 8 (incorrect)

Option 3: 68 = 6+8 = 14 which is not equal to 8 (incorrect)

Option 4: 62 = 6+2 = 8 which is EQUAL to 8 (CORRECT)

Therefore, the missing value is 62 (Option 4)

Learn more:

1. Which of the following options will best complete the relationship given below? 13(4624)33 15(5735)42 11(????)37​

brainly.in/question/11140008

2. 5 7 (37) 8 4(20) 10 6(34) 6 4(?) Find missing number

brainly.in/question/16813585

Answered by Anonymous
0

Step-by-step explanation:

Answer:

Option C is correct Answer

Why?

First of all let's review table of 9:

\begin{gathered}\begin{gathered} \boxed{ \begin{array}{c |c} \tt9 \times 1& \tt9 \\ \tt9 \times 2& \tt \red{18} \\ \tt9 \times 3& \tt \purple{27} \\ \tt9 \times 4& \tt36\\ \tt9 \times 5& \tt \orange{45} \\ \tt9 \times 6& \tt54\\ \tt9 \times 7& \tt63 \\ \tt9 \times 8& \tt72 \\ \tt9 \times 9& \tt81 \\ \tt9 \times 10&\tt90\\ \end{array}} \end{gathered}\end{gathered}

Yep as you can see in 9 table all digits i.e. 18 , 27 and 45 is there,therefore correct option is A.

Now Let's prove why not other options:

Why not option A:

As we know multiples of 5 always end with 5 and 0.

where as in 18 it ends with 8 and in 27 it ends with 7.

Why not option B:

We know that multiples of 6 should also be multiples of both 2 and 3 .

where as 45 is a multiple of 3 but not multiple of 2. .°. it's also not multiple of 6.

Why not option C:

As we know multiple of 12 should also be multiples of both 6 and 3 .

where as 45 is a multiple of 3 but not multiple of 6(we have proved before).°. it's also not multiple of 12.

#CarryOnLearning:D

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