ratio and proportion Question no. 1 The product of the ages of Ankit and Sachin is 240. If twice the age of Sachin is more than Ankit’s age by 4 years, what is Sachin’s age? looks_one 24 looks_two 27 looks_3 40 looks_4 None of these Answer Option looks_4 None of these Solution Solution: let Ankit age is x , Sachin is y According to the Question xy = 240 2y = x+4 ⇒ x =2y - 4 xy = 240 ⇒ (2y - 4)(y) = 240 2y 2 -4y - 240 = 0 ⇒ y 2 -2y -120 = 0 y 2 -12y +10y -120 =0 (y -12)(y + 10) = 0 y = 12 , -10 y =12 (age can't be negative) Question no. 2 Two numbers are in the ratio 3:5. If 9 is subtracted from each of the numbers, their ratio become 12:23. Find the greater of the two numbers. looks_one 55 looks_two 65 looks_3 45 looks_4 None of these Answer Option looks_one 55 Solution Solution : let first number is a , second number is b According to question a/b = 3/5 ⇒ 5a = 3b (a-9)/(...

General Instruction 1. All the questions are compulsory 2. The question paper consists of 40 question and it is divided into four sections A,B , C and D. SECTION A Comprises of 20 questions carrying 1 mark each. Section B comprises of 6 questions carrying 2 marks each, Second C comprises of 8 questions carrying 3 marks cach. Section D comprises of 6 questions carrying 4 marks each. 3. There is no overall choice 4. Use of calculator is not permitted. Section A 1. The LCM of two numbers is 1200. Which of the following cannot be their HCF? A) 4 (b) 3 (c) 5 (d) 6 2. The median of a given frequency distribution is found graphically with the help of (a) histogram (b) frequency curve (c) frequency polygon (d) ogive 3. If the arithmetic mean of x, x+3, x+6,x+9 and x+12 is 10, then x = (a) 1 (b) 2 (c) 6 (d)4 4. Two different dice are tossed together. The probability that the product of two numbers on the top of dice is 6 is a) 1/3 b) 1/6 c) 1/9 (d) ...

Inverse Trigonometry Question no. 31 The value of cot(cosec -1 5/3 + tan -1 2/3) looks_one 4/17 looks_two 5/17 looks_3 6/17 looks_4 3/17 Answer Option looks_3 6/17 Solution Solution : Question no. 32 The Value of is looks_one π/4 looks_two π/2 looks_3 π/3 looks_4 π/6 Answer option looks_4 π/6 Solution Solution : Question no. 33 If x,y ,z are in A.P. and tan -1 x , tan -1 y, tan -1 z are also in A.P. , then : looks_one x=y=z looks_two 2x=3y=6z looks_3 6x=3y=2z looks_4 6x=4y=3z Answer option looks_one x=y=z Solution Solution : x,y ,z are in A.P. 2y = x + z tan -1 x , tan -1 y, tan -1 z are also in A.P. 2tan -1 y = tan -1 x + tan -1 z only x = y = z satisfy it. Question no. 34 Let x ∈(0,1) , the set of all x such that sin -1 x > cos -1 x is the interval looks_one (1/2, 1/√ 2 ) looks_two (1/√ 2 , 1) looks...

Inverse Trigonometry Question no. 16 what is equal to ? looks_one 0 looks_two 2 looks_3 1 looks_4 1/2 Answer Option looks_3 1 Solution Solution : Question no. 17 If sin -1 (1) +sin -1 (4/5)=sin -1 (x) then what is x equal to ? looks_one 3/5 looks_two 4/5 looks_3 1 looks_4 0 Answer Option looks_one 3/5 Solution Solution : Question no. 18 tan -1 (1+x) +tan -1 (1-x)=π/2 is satisfied by looks_one x=1 looks_two x=-1 looks_3 x=0 looks_4 x=1/2 Answer Option looks_3 x=0 Solution Solution : Question no. 19 If , where x= looks_one 1/√ 2 looks_two -1/√ 2 looks_3 ∓ √ (5/2) looks_4 ∓ 1/2 Answer Option looks_4 ∓ 1/2 Solution Solution : Question no. 20 tan -1 1 + tan -1 2 + tan -1 3= looks_one π/2 looks_two π looks_3 π/4 looks_4 none of these Answer Option looks_two ...

Probability Question no. 31 A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads is looks_one looks_two looks_3 looks_4 none of these Answer Option looks_3 Solution Solution : P(x=7) = P(x=9) P(r=2) = 16 C 2 (1/2) 16 = 15/2 13 Question no. 32 A rifleman is firing at a distant target and has only 10% chance of hitting it. The least number of rounds, he must fire in order to have more than 50% chance of hitting it at least one is looks_one 11 looks_two 9 looks_3 7 looks_4 5 Answer Option looks_3 7 Solution Solution : P(hit)= 0.1 P(at least one) = P(x=1) + P(x=2)....+P(x=n) = 1-P(x=0) P(at least one) > 0.5 1-P(x=0) > 0.5 0.5 < P(x=0) 0.5 < (0.9) n n = 7 Question no. 33 If in a binomial distribution n =4 , P(X=0) = 16/81, Then P (X = 4) equals looks_one ...