Home
Class 14
REASONING
sqrtAFI = M : sqrtADD = L ·· sqrtABA =?...

`sqrtAFI `= M : `sqrtADD` = L ·· `sqrtABA` =?

A

I

B

N

C

O

D

K

Text Solution

Verified by Experts

The correct Answer is:
D
`sqrtAFI` = M`sqrtADD `= L
`sqrt144`=12
Similarly,
.`sqrtABA` = .`sqrt121` = 11 = K
Promotional Banner

Topper's Solved these Questions

  • ANALOGY OR SIMILARITY

    KIRAN PUBLICATION|Exercise TYPE-IV (Select the related letter from the given alternatives.)|230 Videos

  • ANALOGY OR SIMILARITY

    KIRAN PUBLICATION|Exercise TYPE-V|37 Videos

  • ANALOGY OR SIMILARITY

    KIRAN PUBLICATION|Exercise TYPE-III|373 Videos

  • ARRANGEMENT OF WORDS IN LOGICAL ORDER

    KIRAN PUBLICATION|Exercise TYPE II|199 Videos

Similar Questions

Explore conceptually related problems

1m^(3) = ? L

If L = 2.01 m +- 0.01 m , what is 3L?

Explain given reasons, which of the following sets of quantum numbers are not possible. {:((a) n = 0 ",", l = 0",", m_l=0",", m_s = + 1//2),((b) n = 1 ",", l = 0",", m_l=0",", m_s = - 1//2),((c) n = 1 ",", l = 1",", m_l=-0",", m_s = + 1//2),((d) n = 2 ",", l = 1",", m_l=0",", m_s = - 1//2),((e) n = 3 ",", l = 3",", m_l=-3",", m_s = + 1//2),((f) n = 3 ",", l = 2",", m_l=0",", m_s = + 1//2):}

In a \ A B C , If L\ a n d\ M are points on A B\ a n d\ A C respectively such that L M B Cdot Prove that: a r\ (\ L C M)=a r\ ( L B M)

ABCD is a square of lengths a, a in N, a gt 1 . Let L_(1),L_(2),L_(3) ... Be points BC such that BL_(1) = L_(1)L_(2) = L_(2)L_(3) = ....=1 and M_(1), M_(2), M_(3)... be points on CD such that CM_(1) = M_(1)M_(2) = M_(2)M_(3) = ...=1 .Then, sum_(n= 1)^(a-1)(AL_(n)^(2) + L_(n)M_(n)^(2)) is equal to

ABCD is a square of length a, a in N, a > 1. Let L_1, L_2 , L_3... be points on BC such that BL_1 = L_1 L_2 = L_2 L_3 = .... 1 and M_1,M_2 , M_3,....be points on CD such that CM_1 = M_1M_2= M_2 M_3=... = 1. Then sum_(n = 1)^(a-1) ((AL_n)^2 + (L_n M_n)^2) is equal to :

l varies directly as m and l is equal to 5, when m=2/3 Find l when m=16/3

Which one of the following is the correct sequence in respect of the Roman numerals: C , D , L a n d M ? C > D > L > M (b) M > L > D > C M > D > C > L (d) L > C > D > M

If l,m,n are real l!=m, then the roots of the equation (l-m)x^(2)-5(l+m)x-2(l-m)=0 are a.real and equal b.Complex c.real and unequal d.none of these