Tutor profile: Sudama S.
A regular hexagon of side 10 cm is inscribed in a circle. Another circle is inscribed in the hexagon. What is the area between the two circles?
Ans: 25* Pi A regular hexagon is inscribed in the circle. So, the radius of circle = side of hexagon Hence, radius for outer circle is R = 10 cm Area of outer circle = Pi * R* R = Pi* 10 *10 = 100 Pi Radius of inner circle r = height of each of equilateral triangle of the hexagon = root(3) * 10 /2 r= 5* root(3) cm Area of inner circle = Pi* r* r = Pi * 5* root (3) * 5* root (3) = 75 Pi Area between the two circles = Area of outer circle- Area of inner Circle Area= 100 Pi -75 Pi = 25 Pi
Subject: Basic Math
A shopkeeper bought 100 articles from a wholesaler at 30% discount.He sold all goods at 23% discount, yet making a profit of 10%.What is the ratio of market price in the shop to the market price in the wholesale market
Ans: 1:1 Lets assume the marked price in wholesale market be X Selling price for wholesaler= X - (30%of X) = X - 0.3X = 0.7X Cost price for shopkeeper will be same as the selling price for the wholesaler Selling price for shopkeeper= Cost price + Profit = 0.7X + ( 10% of 0.7X ) = 0.77X Since, the shopkeeper gave 23% discount on marked price M therefore, M - (23% of M)= Selling price for shopkeeper = 0.77X 0.77M = 0.77X Hence M= X So, the ratio is 1:1
Consider the equation x^2- 35x+k=0. How many values pf K exist such that both the roots of the equation are prime numbers.
Ans: 0 Lets A and B are the roots of the equation. We know that sum of roots of a quadratic equation is -b/a. So for this equation, the sum of roots will be -(-35)/1 which is equal to 35. A+B= 35 such that A and B are prime. Now, since A+B is an odd number, it is possible only if one of A and B is even and the other is odd. 2 is an only even prime number. Hence one number has to be 2. Lets assume A=2, then B=35-2 =33 33 is not a prime number. Hence it is not possible to have both primes So, a number of values of K=0.
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