Question

Answer these questions and I want solution/reason/concept behind it Let's see how many people are going to crack JEE :-)​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{\tt Explanation:-}}}$

$\rule{300}{1.5}$

The applied trick is,

$\large{\boxed{\tt X = (a + b + c +d)^2 - 1}}$

[For the values of "X" , "a", "b" , "c" and "d" refer the attachment]

Now, Applying trick,

For the First case :-

$\large{\tt 168 = (2 + 6 + 3 + 2)^2 - 1}$

$\large{\tt 168 = (8 + 5)^2 - 1}$

$\large{\tt 168 = (13)^2 - 1}$

$\large{\tt 168 = 169 - 1}$

As (13²) = 169.

$\large{\boxed{\tt 168 = 168}}$

Hence verified.

For the Second case :-

$\large{\tt 120 = (3 + 2 + 5 +1 )^2 - 1}$

$\large{\tt 120 = (5 + 6)^2 - 1}$

$\large{\tt 120 = (11)^2 - 1}$

$\large{\tt 120 = 121 - 1}$

As (11²) = 121.

$\large{\boxed{\tt 120 = 120}}$

Hence verified.

For the Third case :-

$\large{\tt z = (2 + 3 + 4 + 5)^2 - 1}$

$\large{\tt z = (5 + 9)^2 - 1}$

$\large{\tt z = (14)^2 - 1}$

$\large{\tt z = 196 - 1}$

As (14²) = 196.

$\huge{\boxed{\boxed{\tt z = 195}}}$

So, the missing number is 195 Therefore, Option- (2) is correct.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

The applied trick is:-

$\large{\boxed{\tt X = \left[ \dfrac{a}{b} \right] \times 2}}$

[For the values of "X" , "a", "b" refer the attachment]

Now, applying the trick,

For the First case:-

$\large{\tt 14 = \left[ \dfrac{84}{12} \right] \times 2}$

$\large{\tt 14 = \left[ \dfrac{\cancel{84}}{\cancel{12}} \right] \times 2}$

$\large{\tt 14 = 7 \times 2}$

$\large{\boxed{\tt 14 = 14}}$

Hence verified.

For the Second case:-

$\large{\tt 18 = \left[ \dfrac{81}{9} \right] \times 2}$

$\large{\tt 18 = \left[ \dfrac{\cancel{81}}{\cancel{9}} \right] \times 2}$

$\large{\tt 18 = 9 \times 2}$

$\large{\boxed{\tt 18 = 18}}$

Hence verified.

For the third case:-

$\large{\tt z = \left[ \dfrac{88}{11} \right] \times 2}$

$\large{\tt z = \left[ \dfrac{\cancel{88}}{\cancel{11}} \right] \times 2}$

$\large{\tt z = 8 \times 2}$

$\huge{\boxed{\boxed{\tt z = 16}}}$

So, the missing number is 16 Therefore, Option- (1) is correct.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

The applied trick is:-

$\large{\boxed{\tt X = \dfrac{a + b }{7}}}$

[For the values of "X" , "a", "b" refer the attachment]

Now, Applying the trick is :-

For the First case:-

$\large{\tt 6 = \dfrac{25 + 17}{7}}$

$\large{\tt 6 = \dfrac{42}{7}}$

$\large{\tt 6 = \dfrac{\cancel{42}}{\cancel{7}}}$

$\large{\boxed{\tt 6 = 6}}$

Hence verified.

For the Second case:-

$\large{\tt 8 = \dfrac{38 + 18}{7}}$

$\large{\tt 8 = \dfrac{56}{7}}$

$\large{\tt 8 = \dfrac{\cancel{56}}{\cancel{7}}}$

$\large{\boxed{\tt 8 = 8}}$

Hence verified.

For the Third case:-

$\large{\tt z = \dfrac{89 + 16}{7}}$

$\large{\tt z = \dfrac{105}{7}}$

$\large{\tt z = \dfrac{\cancel{105}}{\cancel{7}}}$

$\huge{\boxed{\boxed{\tt z = 15}}}$

So, the missing number is 15 Therefore, Option- (2) is correct.

$\rule{300}{1.5}$

Answer 2

Question 66)

Middle value is Sum of square of all diagonal digits -1 ..

[(2+6+3+2)²-1]

[13]² - 1

= 169 -1

= 168

similarly,

(3+5+2+1)² - 1

(11)² -1

121 -1

= 120

Similarly we get,

(2+3+4+5)² -1

(14)² -1

196 - 1

= 195 (Ans)

______________________________

Question (67)

we have to find some relationship between 12-14-84

if we divide 14 by 2 and multiply by 12 , we get, 84.

lets see than it will apply in second or not, because in Matrix series we have to apply same logic in all ..

9-18-81

18 half is 9 and when we multiply it by 9 we get 81.

so, our logic is correct ....

(14/2) × 12 = 84

(18/2) × 9 = 81

similarly ,

(x/2) × 11 = 88

(x/2) = 88/11

(x/2) = 8

x = 16 (Ans)

______________________________

Question (68)

we have to Find relationship between 25-17-6

lets try to add 25 and 17 first ,

we get 42 ,

and its divisible of 6

42/6 = 7 .

lets try in other , 38 - 18 - 8

38 + 18 = 56

56/8 = 7 (Same)

so,

in third now ,

89+16 = 105

105/x = 7 ( we have to get same )

x = 105/7 = 15 (Ans)

(Hope it helps you)