Please give me solution of the following questions...
Chapter 18 -> Quadratic Equations -> LOD II
Q 7... If both the roots of the quadratic equation ax^2 + bx + c=0 lie in the interval (0,3), then a lies in a. (1,3) b. (-1,-3) c. (-sr. root(121)/91, - sq root( 8 ) d. None of these.
Q 23. if a^2 b^2 + c^2 = 1 then which of the following can not be the value of (ab + bc + ca)? a. 0 b. 1/2 c. -1/4 d. -1
Thanks in advance for the help :satisfie:
For 1st Q. Suppose x=1 then the eqn becomes a+b+c =0, from this it is certainly not possible to determine interval of a. Even if we use the determinant eqnwe will have 2 variables in that eqn .So it is not possible to determine.Hence (d)
For 2nd Q. we know (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca or ab+bc+ca=/2
Going through options (a) forab+bc+ca=0 ...this implies (a+b+c)^2=1..which is certainly possible.
similarly going through to other options.....
(d)forab+bc+ca=-1 ...this implies (a+b+c)^2= -1 ..which is certainly not possible..it is only possible if a,b,c are imaginary.Hence the ans
Q. Define a no. K such that it is the squares of the sum of the first M natural numbers.(i.e. K = 1^2 + 2^2+.M^2) where M a. 10 b. 11 c. 12 d. none of these Q. M is a two digit no. which has the property that: The product of the factorials of its digits> sum of the factorials of its digits. How many values of M exist? a. 56 b. 64 c. 63 d. none of these Plz xplain the process. regards Gautam.................................:robot:
1) k=M*(M+1)(2m+1)/6...........sum of squares of m terms now K should be divisible by 4 K=4x M*(M+1)(2M+1)/6=4x M*(M+1)(2M+1)/24=x so, M*(M+1)(2m+1) should be divisible by 24 analyse this 24=8*3 2m+1 should always be odd & one of either M & M+1 should be odd so factor 8 will be contributed by either M or M+1 when contributed by M values=8,16,24,32,40 & 48 when M+1 is multiple of 8 values=7,15,23,31,39,47
total values=6+6=12 2) let the two digit no.is xy x!*y!>x!+y! now this can always possible except when one of x & y is 1 or 0 & when x=y=2 so, numbers not following this are 10,11,12,13..........19,20,21,22,30,31..........90,91 Total= 27 Total two digit no.=90 so, total values=90-27=63 plz. confirm the answers
Q. The series of numbers (1, 1/2, 1/3, 1/4..1/1972) is taken. Now two numbers are taken from this series (the first two) say x,y. Then the operation x+y + x.y is performed to get a consolidated number. The process is repeated. What will be the value of the set after all the numbers are consolidated into one number. a. 1970 b. 1971 c. 1972 d. None of these Plz xplain the process. regards Gautam.................................:robot:
let x =1 , y=1/2 x+y+x*y=2 again x=2, y=1/3 x+y+x*y=3 so, doing same upto 1/1972 we get 1972. Answer C
guys there r two more...a bus left point X for point Y..two hours later a car leftpoint X for point Y and arrived at Y at the same time as the bus..if the car and the bus left simultaneously from the opposite ends X and Y towards each other,they would meet 1.33 hours after the start..how much time did it take the bus to travel from X to Y..a)2h b)4h c)6h d)8hamitabh covered a distance of 96km two hours faster than he had planned to..this he achieved by travelling 1km more every hour than he intended to cover every 1 hour 15 minutes..what was the speed at which amitabh travelled during the journey???a)16km/h b)26km/h c)36km/h d)none of these
1) let the time taken=t distance=x speed of bus=x/t speed of train=x/(t-2) 1.33*x/t+1.33*x/(t-2)=x 1.33/t+1.33/(t-2)=1 3t^2-14t+8=0 t=4h 2) let the planned time =t speed=96/t actual time=t-2 actual speed=96/t-2 1h 15mts=5/4h now, 96/(t-2)-5/4*96/t=1 t=8 actual speed=96/(t-2)=96/6=16 hence a
1) let the time taken=t distance=x speed of bus=x/t speed of train=x/(t-2) 1.33*x/t+1.33*x/(t-2)=x 1.33/t+1.33/(t-2)=1 3t^2-14t+8=0 t=4h 2) let the planned time =t speed=96/t actual time=t-2 actual speed=96/t-2 1h 15mts=5/4h now, 96/(t-2)-5/4*96/t=1 t=8 actual speed=96/(t-2)=96/6=16 hence a
q 19.. if a circle is privided with a measure of 19 degree on centre , is it possible to divide the circle into 360 equal parts?? a>never b>pssbl whn one more measure of 20 deg is given c> always d> pssbl if one more measure of 21 degree is given e> none f these
doctors hav advised remu not to hav more dan 20 chocolates a day...when she went to the market complex to get her daily quota,she found out dat if she buys chocolates from da market complex she would hav to pay Rs3 more for the same no. of chocolates than she would hav spent had she bought them from her uncle Scrooge's shop,getting two sweets less per Rs..she finally decided to get them from uncle scrooge's shop paying only in only in Rs 1 coins... i>How many chocolates did she buy ? a>12 b>9 c>18 d>more data reqd. ii>how much would she hav spent at da market complex ? a>Rs6 b>Rs12 c>Rs9 d>Rs5 N.B.: do not blame me if u think da language is confusing ..it is Arun Sharma's fault... i wrote da problem exact as it was dere..am getting some answers but dey are diff.. plz work dis out ..
q 19.. if a circle is privided with a measure of 19 degree on centre , is it possible to divide the circle into 360 equal parts?? a>never b>pssbl whn one more measure of 20 deg is given c> always d> pssbl if one more measure of 21 degree is given e> none f these
:grab:
it is possible always.. on dividing da circle into 19degree parts,,da last part is equal to 18degrees..da diff gives us a one degree part which can be used to divide da circle into 360 equal parts...
ans. S =1*2+2*2^2+3.2^2....................+100*2^100 2S = 1*2^2+2*2^3+......................+99*2^100+100*2^101 substracting we get -S=2+2^2+2^3.................................-100*2^101 S=100*2^101-2(2^100-1)/(2-1) =100*2^101-2^101+2 =99*2^101+2 hence c
The remainder when 10^10 + 10^100+10^1000+...+10^1000000000 is divided by 7 is (a) 0 (b) 1 (c)2 (d) 5
euler no of 7=6 10 & 7 r coprimes so, 10^6 will give remainder 1 when divided by 7 10^n can be written in the form 6K+4 so, each term can be written as 10^4 hence no. will reduce to 9*10^4 9*10^4mod7=2*3*3*3*3mod7=54mod7=5
euler no of 7=6 10 & 7 r coprimes so, 10^6 will give remainder 1 when divided by 7 10^n can be written in the form 6K+4 so, each term can be written as 10^4 hence no. will reduce to 9*10^4 9*10^4mod7=2*3*3*3*3mod7=54mod7=5
Cud u plz tell in detail of Euler theorem or give me any web link stating this theorem. How u say "euler no of 7=6" ? What does it mean ?
If u have any soft copy of this theorem with u, then plz upload it regards
Cud u plz tell in detail of Euler theorem or give me any web link stating this theorem. How u say "euler no of 7=6" ? What does it mean ? If u have any soft copy of this theorem with u, then plz upload it regards
A certain no. of trucks were required to transport 60 tons of steel wire from A to B. However, it was found that since each truck could take 0.5 tons of cargo less, another four trucks were needed. How many trucks were initially planned to be used?a. 10 b. 15 c. 20 d. 25Ans. CQ.) One colective farm got an average harvest of 21 tons of wheat and another collective farm that had 21 acres of land less given to wheat, got 25 ton more from a hectare. As a result, the second farm harvested 300 tons of wheat more than the first. How many tons of wheat did each farm harvest?a. 3150,3450 b. 3250,3550 c. 2150,2450 d. None of theseAns. Aregards GAUTAM.....
