Answer: B x^{2} -3x +2 = (x-2)(x-1) => The factors of the cubic equation is x =2 and x= 1 On putting x=2 , we get, 8-24+2p+q =0 => 2p+q =16........(i) On putting x=1, we get, 1-6+p+q = 0 => p+q =5........(ii) Thus on solving these two equation we get p =11 and q = -6 So p+q >0 and pq <0

Q. No. 2:

The sequences x_{1}, x_{2}.... and y_{1}, y_{2}........ are in arithmetic progressions such that x_{1}+y_{1} = 100 and x_{22}-x_{21} = y_{99} - y_{100}. Find the sum of the first 100 terms of the progression, (x_{1}+y_{1}), (x_{2}+y_{2})...........

Answer: C The condition x_{22}-x_{21} = y_{99} - y_{100}. implies that the common difference of the two arithmetic progressions are the negatives of each other. Therefore, the series (x_{1}+y_{1}),(x_{2}+y_{2}), is a constant series , in which each term is equal to (x_{1}+y_{1}) =100 Thus Required Sum = 100 * 100 = 10,000

P divides his property among his four sons A, B,C and D after donating Rs 20,000 and 10% of his remaining property. The amounts received by the last three sons are in A.P and the amount received by the fourth son is equal to the total amount donated. The first son receives as his share Rs 20,000 more than the share of second son. The last son received Rs 1 lakh less than the eldest son.

Answer: D All amounts are in thousands of rupees. Lets P's total property be (20+10p) The donation is 20+p The 4th son's share is 20+p The 1st son's share is 120+p The 2nd son's share is 100+p The 3rd son's share is 60+p As, 4th , 3rd and 2nd are in AP Thus, 300+4p = 9p => p= 60. P's total donation is 20+p = 80 thousand.

Answer: C All amounts are in thousands of rupees.
Lets P's total property be (20+10p)
The donation is 20+p
The 4th son's share is 20+p
The 1st son's share is 120+p
The 2nd son's share is 100+p
The 3rd son's share is 60+p
As, 4th , 3rd and 2nd are in AP
Thus, 300+4p = 9p
=> p= 60. The 3rd son's share is 60+p = 120 thousands.

Q. No. 4:

When the index of an exponential expression with a positive base is doubled, then the expression increases by 700%. If one of the values that the base can not have is X which of the following is not a possible value of P?

Answer: D Let the expression be a^{n}. => a^{2n} = 8a^{n} => n = log_{a} 8. a can be 5, 4 and 8 but it cannot be 1.

Q. No. 5:

Abhishek had a certain number of Re1 coins, Rs 2 coins and Rs 10 coins. If the number of Re 1 coins he had is six times the number of Rs 2 coins Abhishek had, and the total worth of his coins is Rs 160, find the maximum number of Rs 10 coins Abhishek could have had.

Answer: A If the Abhishek had x Re 1, y Rs 2 coins and z Rs 10 coins, the total value of coins he had = x(1)+y(2)+z(10) = x+2y+10z = 160. Since, 6y =x Thus , 8y + 10z =160 i.e 8y is a multiple of 10 i.e y=5 or y=10 i.e (x,y,z) = (30,5,12) or (60,10,18) Thus, the maximum value of 'z' is 12.

Q. No. 6:

In an A.P, the 12th term is 7 times the 2nd term and the 8th term is 3 more than 10 times the first term. What is the 5th term of the G.P whose first term is the first term of A.P and whose common ratio is equal to the common difference of the A.P.

Answer: A Let the progression be a, a+d, a+2d.......... => a+11d = 7a +7d => 6a = 4d => 3a =2d Also, a+7d = 10a+3 => 7d = 9a+3 => 7d = 6d+3 => d=3, a=2 The GP is 2, 2(3), 2(9),2(27), 2(81) The firth term is 162.