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Answers: Answers 1. Let the actual number of people be “x” So, we will rephrase the question: There is provision for (60 – 15 = 45 days) for “x” people. The provisions last 40 days if the number of people becomes “x + 500” Using chain rule : 40 * (x + 500)/x =45 Or, 40x +20000 = 45x Or, 5x = 20000 Or, x = 4000 answer
2. Initial distribution of money = 7:6:5. Say, the total amount was Rs 180 (i.e. sum of ratios) So, Rohan had 70, Rakesh had 60 and Ramesh 50. After the game of cards, the ratio becomes 6:5:4 So, now each one has Rohan = (6/15) x 180 =72, Rakesh = (5/15) x 180 = 60, Ramesh = (4/12)x 180 = 48 So, Rohan has won 72 – 70 = 2. But in actual Rohan has won Rs 1200. Actual amount with Rohan = 70 x 1200/2 = 42000
3. First no. is T1. T2 = T1 + 2, T3 = T2 + 2 = T1 + 2 x 2 T4 = T3 + 2 = T1 + 3 x 2 .... T50th number is t1 + 49 x 2 S = T1 + (T1 + 2) + (T1 + 4) +......... + (T1 + 98) = 50T1 + 2 (1 + 2 + ....49) S = 50T1 + 2 x 49 x 50/2 = 50T1 + 2450 But S = 3250 So, T1 =(3250 – 2450)/50 = 16 T50 = 16 + 49 x 2 = 114 So,T1 + T50 = 16 + 114 = 130
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1. Let the actual number of people be “x”
So, we will rephrase the question:
There is provision for (60 – 15 = 45 days) for “x” people. The provisions last 40 days if the number of people becomes “x + 500”
Using chain rule :
40 * (x + 500)/x =45
Or, 40x +20000 = 45x
Or, 5x = 20000
Or, x = 4000 answer
2. Initial distribution of money = 7:6:5.
Say, the total amount was Rs 180 (i.e. sum of ratios)
So, Rohan had 70, Rakesh had 60 and Ramesh 50.
After the game of cards, the ratio becomes 6:5:4
So, now each one has
Rohan = (6/15) x 180 =72, Rakesh = (5/15) x 180 = 60, Ramesh = (4/12)x 180 = 48
So, Rohan has won 72 – 70 = 2.
But in actual Rohan has won Rs 1200.
Actual amount with Rohan = 70 x 1200/2 = 42000
3. First no. is T1.
T2 = T1 + 2,
T3 = T2 + 2 = T1 + 2 x 2
T4 = T3 + 2 = T1 + 3 x 2
....
T50th number is t1 + 49 x 2
S = T1 + (T1 + 2) + (T1 + 4) +......... + (T1 + 98) = 50T1 + 2 (1 + 2 + ....49)
S = 50T1 + 2 x 49 x 50/2 = 50T1 + 2450
But S = 3250
So, T1 =(3250 – 2450)/50 = 16
T50 = 16 + 49 x 2 = 114
So,T1 + T50 = 16 + 114 = 130
13. Since, X3 – 6x2 + px + q is exactly divisible (x2 – 3x + 2), let (x + a) be the quotient.
ReplyDeleteTherefore, (x2 – 3x + 2) (x + a) = X3 + (a – 3) x2 + (2 – 3a) x + 2a = X3 – 6x2 + px + q
Comparing the coefficients we get
(a – 3) = -6, or a = -3
2 – 3a = p, or p = 2 – 3*-3 = 11
2a = q, or q = 2* -3 = -6
P + q = 11 – 6 = 5, or p + q > 0
Pq = 11 * -6 = -66 or pq < 0