AIEEE Questions

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The skul contains the bones like Occipital bone, Frontal bone, Mandible, Temporal bone, Parietal bone. The mandible, or jaw bone, is the only movable bone of the skull, forming the temporomandibular joint with the temporal bone. The lower teeth are rooted into the mandible while the upper teeth are rooted in the two maxillae.

PSRQ --> {16, 19, 18, 17} - next 3 alphabets of "P" is written in reverse order
SVUT --> {19, 22, 21, 20} - next 3 alphabets of "S" is written in reverse order

MNPO --> {13, 14, 16, 15}

KNML --> {11, 14, 13, 12} - next 3 alphabets of "K" is written in reverse order

Except "MNPO", others are not in consecutive order.

4 is father of 1 and grandfather of 6. So, 1 is father of 6. Thus, 4 and 1 are the two fathers.
3 is the sister of 6. So, 3 is the daughter of 1. Since there is only one mother, it is evident that 5 is the wife of 1 and hence the mother of  3 and 6. So, 2 is brother of 1. There are three brothers. So, 6 is the brother of 3. Hence, 2 and 1 are brothers. 6 is the brother of 3. So, three brothers are 1, 2, 6.

PARAM 8000 was was the first supercomputer developed by Indian scientists launched on July 01, 1991 is considered India's first supercomputer.

• It was indigenously built in 1991 by Centre for Development of Advanced Computing and was replicated and installed at ICAD Moscow in 1991 under Russian collaboration.

The mutiplicative inverse of a number is nothing but a reciprocal of a number.

Now, the product of a number and its multiplicative inverse is always equal to 1.

For example :

Let the number be 15

Multiplicative inverse of 15 = 1/15

The product of a number and its multiplicative inverse is = 15 x 1/15 = 1.

FCH₂COOH is the strongest acid out of all.

Since the electron withdrawing inductive effect (-I effect) of the halogens decreases in the order of F>Cl> Br> I , therefore the acidic strength of the α- halo acids decreases in the same order.

FCH₂COOH  >  ClCH₂COOH   >   BrCH₂COOH   >   ICH₂COOH.

Ctrl key is to be pressed To copy cell contents using drag and drop.

Now, Total work = LCM(16, 8) = 48

A's one day work = + 48/16 = + 3

B's one day work = - 48/8 = -6

Given A worked for 5 days to build the wall => 5 days work = 5 x 3 = + 15

2days B joined with A in working = 2(3 - 6) = - 6

Remaining Work of building wall = 48 - (15 - 6) = 39

Now this remaining work will be done by A in = 39/3 = 13 days.