According to CIP rules the configuration of chiral carbon atoms in
D- (+)-Glucose are
(A)2S, 3S, 4R, 5S
(B)2S, 3S, 4R, 5R
(C)2R, 3R, 4S, 5S
(D) 2R, 3S, 4R, 5R 

Which of the following sugars has the configuration (2S 3R, 4R) ?

If R = (x^(2)+2)/(x-3) and S =(x-1)/x , then find (a) R + S (b) R - S (c) R.S (d) R/S

Divide Rs. .671 among A, B and C such that if their shares be increased Rs.3, Rs. 7 and Rs. 9 respectively the remainder shall be in the ratio 1 :2 : 3. R s .110 , R s .220 , R s .336 b. R s 112 , R s .223 , R s .336 c. R s .105 , R s .223 , R s .330 d. none of these

A: Mn shows exceptional M.P. in 3d series R: Its outer configuration is 4s^(2) 3d^(5)

If S_r denotes the sum of the first r terms of an A.P. Then, S_(3n)\ :(S_(2n)-S_n) is n (b) 3n (c) 3 (d) none of these

S_(r) denotes the sum of the first r terms of an AP.Then S_(3d):(S_(2n)-S_(n)) is -

Anu is a working partner and I3imla is a sleeping partner in a business. Anu puts in Rs. 5000 and Bimla puts in Rs. 6000. Anu receives 12.5% of the profit for managing the business and the rest is divided in proportion to their capitals. What does of a profit of Rs. 880? R s .400 , R s .480 b. R s .450 , R s .430 c. R s .460 , R s .420 d. R s .470 , R s .410

(XAT(r, s))/(MAT(r,s)) = 2 (i) r = 3, s = 5 (ii) r = -3, s = -5

The following are the steps in finding the matrix B, if B + ({:(2, 3), (4, 5):}) = ({:(5, 4), (3, 2):}) . Arrange them in sequential order. (A) therefore ({:(p, q), (r, s):}) + ({:(2, 3), (4, 5):}) = ({:(5, 4), (3, 2):}) (B) Let B = ({:(p, q), (r, s):}) (C) ({:(p+2, q+3), (r+4, s+5):}) = ({:(5, 4), (3, 2):}) (D) P + 2 = 5, q + 3 = 4, r+ 4 = 3, s + 5 = 2 rArr p = 3, q = 1, r = -1, s = -3 (E) therefore B = ({:(3, 1), (-1, -3):})

If P: Q = 2 : 3 , Q : R = 5 : 7 and R : S = 3 : 10 then P : S = ?