In Arun Sharma's Number System Chapter,there is a Topic related to Finding the highest power (n) of a number which divides a Factorial number(xy00!) completely. In this topic, i have a major problem in solving the different type of questions.
Ques1.Highest Power of 82 which divides 342! Ans1. 7x3x2x2 . In this case which check only for no. of 7s.
But in Ques2. Highest Power of 175 which divides 344! Ans2. 5x5x7 . But in this case we find no. of 7s as well as 5^2s.
Why so?
Also provide solutions to the below Questions with Explanations:-
Ques1> Highest Power of 12600 which divides 50! Ques2>Highest Power of 720 which divides 77! Ques3> Highest Power of 2520 which divides 50!
P.S. The main point of concern is how we determine which terms need to be counted for finding the answer?
Dude these all ques are solved by a specific method ques 1 > soln > prime factorise 12600= 2^3 x 3^2 x 5^2 x 7 then u find the maximum power of the prime factrs => 2,3,5,7 (u found them above) in 50!
now max. power of 2 in 50! = (50/2) + (50/4) + (50/ + (50/16) + (50/32) = 47 , Now divide this 47 by 3 {47/3 = 15} as 2 was raised to power 3 in the prime factorisation of 12600 , so now max. power of 2 in 50! = 15
max. power of 3 in 50! = (50/3) + (50/9) +(50/27) = 22 , now divide this 22 by 2 {22/2=11} as 3 was raised to power 2 in prime factorisation of 12600 so now max. power of 3 in 50! = 11
similarly max. power of 5 in 50! = 6
similarly maximum power of 7 in 50! = 8
the final answer will be the least power of prime factors found which is 6 highest power of 12600 that perfectly divides 50! = 6
Pg 166 Q 26: In a village consisting of p persons, x% can read and write. Of the males alone y% , and of the females alone z% can read and write. Find the number of males in the village in terms of p,x,y and z if z
@Pg 166 let total No. of men be m, total number of persons who can read and write = xp, xp=ym + z(p-m) m=p(x-y)/(y-z) Ans :- (d)
Circles are drawn with the four vertices as the center and radius equal to the side of a square. If the square is formed by joining the mid-points of another square of side 2 6 , find the area common to the four circles.
Q2. Geometry and Mensuration Preassessment test, question nos 11
Geometry Pre-Assess Test Q.11
In the figure given below, a circle is inscribed inside a square. In the gap between the circle and the square (at the corner) a rectangle measuring 20cm X 10cm is drawn such that the corner A of the rectangle is also a point on the circumference of the circle. What is the radius of the circle in cm? a) 30cm b) 40cm c)50cm d)None of these
chapter 10 (time speed and distance) level 1 ques 66 A man can row 30 km upstream and 44 km downstream in 10 hours. It is also known that he can row 40 km upstream and 55 km downstream in 13 hours. Find the speed of the man in still water. a) 4 km/h b) 6 km/h c) 8 km/h d)12 km/h
chapter 10 (time speed and distance) level 1 ques 66 A man can row 30 km upstream and 44 km downstream in 10 hours. It is also known that he can row 40 km upstream and 55 km downstream in 13 hours. Find the speed of the man in still water. a) 4 km/h b) 6 km/h c) 8 km/h d)12 km/h
Please help me with a solution.
Let time taken for upstream row in first case = t1 and that in case 2 is t2.
As, the up/down stream speed should be equal, we can write,
chapter 10 (time speed and distance) level 1 ques 66 A man can row 30 km upstream and 44 km downstream in 10 hours. It is also known that he can row 40 km upstream and 55 km downstream in 13 hours. Find the speed of the man in still water. a) 4 km/h b) 6 km/h c) 8 km/h d)12 km/h
Please help me with a solution.
i am not telling u the correct way ... but if u find it always tough
chapter 10 (time speed and distance) level 1 ques 66 A man can row 30 km upstream and 44 km downstream in 10 hours. It is also known that he can row 40 km upstream and 55 km downstream in 13 hours. Find the speed of the man in still water. a) 4 km/h b) 6 km/h c) 8 km/h d)12 km/h
Please help me with a solution.
let v=speed of still water u-speed of stream
then v-u=speed upstream v+u=speed downstream
30/v-u + 44/v+u =40 40/v-u + 55/v+u = 13 solving the above two equation gives v+u=11 and v-u=5 Hence, the v=speed of still water = 8 was an easy one
In ABC, D is the midpoint of BC. E is a point on AC such that AE : EC = 2 : 1 and F is a point on AB such that AF : FB = 3 : 1. Line segments AD and FE intersect at point O. What is the ratio of the area of DOF to the area of DOE?
can nybdy explain the following problem with proper soln,,,
N=202*20002*200000002*200...2(15zeros) * 2000.......2(31zeros). sum of digit in this multiplication will be : 112; 160; 144; can't be determined??
############common u people ...solve the problem yar.. Estaller whr are you... waiting 4 the past 3-4 days to get the damn soln..of above problem..
N= 2*101*2*10001*2*100....1(15 zeros)*2*100...1(31 zeros) N= 16*(101)*(10001)*(1000...1 (15zeros ))*(10000..1(31 zeros)) N= 16*(10^2 +1)(10^4 + 1)(10^16 +1)(10^32 +1) on solving above you will get 16 times 16 in N
In ABC, D is the midpoint of BC. E is a point on AC such that AE : EC = 2 : 1 and F is a point on AB such that AF : FB = 3 : 1. Line segments AD and FE intersect at point O. What is the ratio of the area of DOF to the area of DOE?
Options: 5:4 9:8 8:9 4:3
Let the area(ABC) = 2x
Then ar(ABD) = ar(ACD) = x ar(ADF) = 3x/4 and ar(BDF) = x/4 ar(ADE) = 2x/3 and ar(CDE) = x/3
+ (50/16) + (50/32) = 47 