Here, you will find all the questions of ISI Entrance Paper 2017 from Indian Statistical Institute's B.Stat Entrance. You will also get the solutions soon of all the previous year problems. Problem 1 : Let the sequence \( \{ a_n\} _{n \ge 1 } \) be defined by $$ a_n = \tan n \theta $$ […]
Here are the problems and their corresponding solutions from BStat Hons Objective Admission Test 2005. Try it yourself and then read the solutions.
Here, you will find all the questions of ISI Entrance Paper 2016 from Indian Statistical Institute's B.Stat Entrance. You will also get the solutions soon of all the previous year problems. Problem 1: In a sports tournament of $n$ players, each pair of players plays exactly one match against each other. There are no draws. […]
Try this problem from TOMATO Objective 288, useful for ISI BStat, BMath Entrance Exam based on finding big remainder in a small way. Problem: Tomato objective 288 The remainder R(x) obtained by dividing the polynomial [latex]x^{100}[/latex] by the polynomial [latex]x^2-3x+2[/latex] is (A) [latex]2^{100}-1[/latex] (B) [latex](2^{100}-1)x-(2^{99}-1)[/latex] (C) [latex]2^{100}x-3(2^{100})[/latex] (D) [latex](2^{100}-1)x+(2^{99}-1)[/latex] SOLUTION: (B) The the divisor is […]
Problem: If the roots of the equation ${(x-a)(x-b)}$+${(x-b)(x-c)}$+${(x-c)(x-a)}$=$0$, (where a,b,c are real numbers) are equal , then (A) $b^2-4ac=0$ (B) $a=b=c$ (C) a+b+c=0 (D) none of foregoing statements is correct Answer: $(B)$ ${(x-a)(x-b)}$+${(x-b)(x-c)}$+${(x-c)(x-a)}$=$0$ => $x^2-{(a+b)}x$+$ab+x^2-{(b+c)}x$+$bc+x^2-{(c+a)}x+ca$=$0$ => $3x^2-2{(a+b+c)}x$+$(ab+bc+ca)$=$0$ discriminant, of the equation is => $4{(a+b+c)^2}$-$4.3{(ab+bc+ca)}$=$0$ => $a^2+b^2+c^2+2(ab+bc+ca)$-$3(ab+bc+ca)$=$0$ => $a^2+b^2+c^2$-$(ab+bc+ca)$=$0$ => $a=b=c$ So, option (B) is correct.
Try this beautiful problem from TOMATO Objective no. 258 based on Real Roots of a Cubic Polynomial. Problem: Real Roots of a Cubic Polynomial Let a,b,c be distinct real numbers. Then the number of real solution of [latex](x-a)^3+(x-b)^3+(x-c)^3=0[/latex] is (A) 1 (B) 2 (C) 3 (D) depends on a,b,c Solution: Ans: (A) Let [latex]f(x)=(x-a)^3+(x-b)^3+(x-c)^3[/latex] [latex]=> f'(x)=3(x-a)^2+3(x-b)^2+3(x-c)^2=0[/latex] [latex]=> […]
Try this beautiful problem from TOMATO Objective no. 257 based on Roots of a Quintic Polynomial. Problem: Roots of a Quintic Polynomial The number of real roots of [latex] x^5+2x^3+x^2+2=0[/latex] is (A) 0 (B) 3 (C) 5 (D) 1 Solution: Answer: (D) [latex] x^5+2x^3+x^2+2=0[/latex] [latex] \implies x^3(x^2+2)+(x^2+2)=0[/latex] [latex] \implies (x^3+1)(x^2+2)=0[/latex] [latex] \implies (x+1)\bold{\underline{(x^2-x+1)(x^2+2)}}=0[/latex] The expression in underline doesn't have any […]
Problem: The number of terms in the expression of $latex [(a+3b)^2 (a-3b)^2]^2 $ A) 4; B) 5; C) 6; D) 7; Solution: $latex [(a+3b)^2 (a-3b)^2]^2 $ $latex = [\{(a+3b)(a-3b)\}^2]^2 $ $latex = \{ (a^2 -9b^2)^2\}^2 = (a^2 - 9b^2)^4 $ By Binomial Theorem, the given expression contains 5 terms (since $latex (x +y)^n $ has […]
Try this beautiful problem from TOMATO Objective no. 27 based on Closure of a set of even numbers. Problem: Closure of a set of even numbers S is the set whose elements are zero and all even integers, positive and negative. Consider the 5 operations- [1] addition; [2] subtraction; [3] multiplication; [4] division; and […]
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Complex numbers and Sets.
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Consecutive Positive Integers.
Try this beautiful problem from algebra, based on Sum of the numbers from AMC-10A, 2001. You may use sequential hints to solve the problem.
Try this beautiful problem from Geometry based on pentagon and square pattern from AMC-10A, 2001. You may use sequential hints to solve the problem.
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on combinatorics in Tournament.
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Fair Coin Problem.
Try this beautiful problem from algebra, based on algebraic equations from AMC-10A, 2001. You may use sequential hints to solve the problem.
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Ordered pair. You may use sequential hints.
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on combinatorics in Tournament.
Try this beautiful problem from Geometry: Area of region from AMC-10A, 2007, Problem-24. You may use sequential hints to solve the problem.