Q1. 17n - 31n + 33n - 19n is definitely a multiple of which of the following numbers, given that n is an odd?
Q2. The infinite sum 1 + 4/7 + 9/49 + 16/343 + 25/2401 + . . . . equals
Q3. The number of roots common between the two equations x3 + 2x2 + 7x + 3 =0 and x3 + 3x2 + 4x + 5 =0 is
Q4. Coefficient of x16 is the expansion of ( x2- 2x )10 is :
Q5. Sachin took 1 wicket for 15 runs in his first test match, 2 wickets for 25 runs in the next match, 3 for 35 in the third match and so on. Calculate his runs per wicket after 8 matches. Also, he made 4 run outs in 8 matches which are counted as wickets earned by him
Q6. A trader mixes 15 kgs of a substance costing Rs.30 , 14kgs of a mixture costing Rs. 28, 13 kgs of a mixture costing Rs.26 and so on, upto 1 kg of a substance costing Rs.2 and sells it at Rs. 2480 per 100kgs. Calculate his percentage profit. __________________
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Humraj
Q1. 17n - 31n + 33n - 19n is definitely a multiple of which of the following numbers, given that n is an odd?
a. 112 b. 180 c. 350 d. None of these
Sol.
=(33n - 19n) - (31n - 17n)
=(33 - 19)(33n-1 +33n-2 191 +....+19n-1) - (31 - 17)(31n-1 +31n-2 171 +....+17n-1)
=14{(33n-1 +33n-2 191....+19n-1) - (31n-1 +31n-2 171....+17n-1)}
but {(33n-1 +33n-2 191....+19n-1) - (31n-1 +31n-2 171....+17n-1)} is multiple of 8.
then (33n - 19n) - (31n - 17n) is multiple of 112.
Short cut: taking n=3 and geting ans.
answers:
Q1) 112
Q2) 211/13
Q3) 2 ( X=1,2)
Q4) None of these (4260)
Q2. The infinite sum 1 + 4/7 + 9/49 + 16/343 + 25/2401 + . . . . equals
a. 27/14 b. 21/13 c. 49/27 d. 256/147
Sol.
S = 1 + 4/7 + 9/49 + 16/343 + 25/2401 + . . . . (i)
S/7 = 1/7 + 4/49 + 9/343 + 16/2401+..........(ii)
(i) - (ii)
6S/7 = 1 + 3/7 + 5/49 + 7/343 + 9/2401+............(iii)
6S/49= 1/7 + 3/49 + 5/343 + 7/2401 + ..........(iv)
(iii) - (iv)
36S/49 = 1 + 2/7 + 2/49 + 2/343 + ............
36S/49 = 1 + 2(1/7 + 1/49 + 1/343...............)
36S/49 = 4/3
S = (4/3)(49/36)
S = 49/27
Q3. The number of roots common between the two equations x3 + 2x2 + 7x + 3 =0 and x3 + 3x2 + 4x + 5 =0 is
a. 0 b. 1 c. 2 d. 3
Sol.
I think no common roots.
Q4. Coefficient of x16 is the expansion of ( x2- 2x )10 is :
a. 3360 b. -3360 c. 980 d. -980 e. None of these
SOL.
The independent term of (x + a)n
Tr+1 = nCr.xn-r.ar
The independent term of (x2 - 2x)10
Tr+1= 10Cr.(x2)10-r.(-2x)r
Tr+1= 10Cr.x20-r(-2)r
therefore, x20-r = x16; r = 4
Coefficient of x16 = 10C4.(-2)4 = 3360
Q6. A trader mixes 15 kgs of a substance costing Rs.30 , 14kgs of a mixture costing Rs. 28, 13 kgs of a mixture costing Rs.26 and so on, upto 1 kg of a substance costing Rs.2 and sells it at Rs. 2480 per 100kgs. Calculate his percentage profit.
a. 15% b. No profit no loss c. 20% d. None of these
Sol.
Total cost price = 15x30 + 14x28 + 13x26.........................+1x2
= 2(152 + 142 + 132 ..........................+12)
= 2.15.16.31/6 = 2480
Total volume = 1+2+3.............+15 = 64kg
Cost price per kg = 2480/64 = Rs. 38.75
Selling price per kg = Rs. 24.80
not porfit; only loss
Thanks a lot humraj
n/a
wickets taken
= 1 + 2 + 3 . . . .
= (8 x 9)/2 = 36 and 4 from run out
So total wickets = 40
Run scored = 15 + 25 + 35. . . . .8 terms
= 8/2 [ 30 + 7 x 10]
= 4 x 100 = 400
So Run/Wicket = 400/40 = 10
n/a
@ Praveen
welcome............
what is the ans of Q. 3 and Q.6?????????
I have doubts bout the solution u gave for this question:
Total cost price = 15x30 + 14x28 + 13x26.........................+1x2
= 2(152 + 142 + 132 ..........................+12)
= 2.15.16.31/6 = 2480
till this point its fine....
Total volume = 1+2+3.............+15 = 64kg -------this makes upto 120 kg
so he mixes 120 kgs at a prices of 2480 and sells 100 kg at a price of 2480.
hence he has a profit which is given by
% profit = (120-100)/100 *100 = 20%
hence the option (c) is correct
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