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Practice question on permutation and combination 31-45 (Set 3)

permutation and combination Question no. 31 In a plane , there are 10 points out of which 4 are collinear , then the number of triangles that can be formed by joining these points are looks_one 60 looks_two 116 looks_3 120 looks_4 None of these Answer Option looks_two 116 Solution Solution : No. of triangles = n C 3 - p C 3 = 10 C 3 - 4 C 3 = 120 - 4 = 116 Question no. 32 The number of straight lines joining 8 points on a circle is looks_one 8 looks_two 16 looks_3 24 looks_4 28 Answer Option looks_4 28 Solution Solution : No. of straight line = n(n-1)/2 = 8 *7 / 2 = 28 Question no. 33 The number of diagonals in a polygon of m sides is looks_one m(m-5)/2! looks_two m(m-1)/2! looks_3 m(m-3)/2! looks_4 m(m-2)/2! Answer Option looks_3 m(m-3)/2! Solution Solution : The number of diagonals in a polygon of m sides is m(m-3)/2! Question no. 34 How many ...

Practice Question on permutation and combination 16-30 (For IIT JEE , NDA , BITS ) Set 2

Permutation and Combination Question no. 16 What is the number of three digit odd numbers formed by using the digits 1,2,3,4,5,6 if repetition of digits is allowed ? looks_one 60 looks_two 108 looks_3 120 looks_4 216 Answer Option looks_two 108 Solution Solution : Number of digits at one's place = 1 , 3 , 5 [Total number is 3] Number of digits at ten's place = 6 [1,2,3,4,5,6] Number of digits at hundred place =6 [ 1, 2 , 3 , 4 , 5 , 6 ] Total number of ways = 3 * 6 * 6 = 108 Question no. 17 What is the number of ways of arranging the letters of the word "BANANA" so that no two N's appear together? looks_one 60 looks_two 40 looks_3 80 looks_4 100 Answer Option looks_two 40 Solution Solution : Total no. of ways = 6!/(3!2!) = 60 Two N's come together = 4!*5 / 3! = 20 No. of ways no two N's appear together = 60 - 20 = 40 Question no. 18 How many times does the ...

Practice Question On LCM and HCF for Bank , SSC and UPSC

Work And time Question no. 1 The H.C.F and L.C.M of two numbers are 21 and 4641 respectively. If one of the numbers lies between 200 and 300, then the two numbers are looks_one 273 , 357 looks_two 273 , 361 looks_3 273 , 359 looks_4 273 , 363 Answer Option looks_one 273 , 357 Solution Solution : HCF = 21 , LCM = 4641 = 13 * 3 * 7 * 17 Each number has 21 as factor First number = 21 * 13 = 273 Second Number = 21 * 17 = 357 Question no. 2 What is the smallest number which when increased by 5 is completely divisible by 8, 11 and 24? looks_one 264 looks_two 259 looks_3 269 looks_4 None of these Answer Option looks_two 259 Solution Solution : LCM ( 8 ,11 , 24) = 264 LCM is the smallest number that is divisible by all their factors 264 is the number completely divisible by 8, 11 and 24 Number is increased by 5 = 264 - 5 = 259 Question no. 3 Find the greatest number that will divide 148,246 and 623 le...

practice question on Trigonometry

Trigonometry Question no. 1 The minimum value of 2sin 2 θ + 3cos 2 θ is looks_one 3 looks_two 2 looks_3 0 looks_4 1 Answer Option looks_two 2 Solution Solution : 2sin 2 θ + 3cos 2 θ = 2sin 2 θ + 2cos 2 θ + cos 2 θ 2(sin 2 θ + cos 2 θ) + cos 2 θ 2 + cos 2 θ -1 ≤ cosθ ≤ 1 0 ≤ cos 2 θ ≤ 1 2 ≤ 2 + cos 2 θ ≤ 3 Question no. 2 The simplest value of sin 2 x + 2tan 2 x–2sec 2 x + cos 2 x is looks_one -1 looks_two 1 looks_3 0 looks_4 2 Answer Option looks_one -1 Solution Solution : sin 2 x + 2tan 2 x–2sec 2 x + cos 2 x sin 2 x + cos 2 x + 2 (tan 2 x - sec 2 x ) = 1 + 2 (-1) = 1 - 2 = -1 Question no. 3 If cos 4 θ– sin 4 θ = 2/3 , then the value of 2cos 2 θ–1 is looks_one 0 looks_two 1 looks_3 2/3 looks_4 3/2 Answer option looks_3 2/3 Solution Solution : Question no. 4 The value of tan4°. tan43°. tan47°. tan86° is looks_one ...

Practice Question On 3 D For IIT-JEE , NDA and AIRFORCE

3-D Question no. 1 Co-ordinate of a point equidistant from the points (0,0,0) ,(a,0,0) , (0,b,0) , (0,0,c) is looks_one looks_two looks_3 looks_4 (a,b,c) Answer Option looks_3 Solution Solution : is the point which is equidistant from the (0,0,0) ,(a,0,0) , (0,b,0) , (0,0,c). Question no. 2 Let (3,4,-1) and (-1 ,2 ,3 ) are the end points of a diameter of sphere. then the radius of the sphere is equal to looks_one 1 looks_two 2 looks_3 3 looks_4 9 Answer Option looks_3 3 Solution Solution : where x 1 , y 1 , z 1 and x 2 , y 2 , z 2 are the end points of diameters. Now, put the value in the formula . Question no. 3 The equation of the sphere touching the three co-ordinate planes is looks_one x 2 + y 2 + z 2 + 2a( x + y + z) + 2a 2 = 0 looks_two x 2 + y 2 + z 2 - 2a( x + y + z) + 2a 2 = 0 looks_3 x 2 + y 2 + z 2 ± 2a( x + y + z) + 2a 2 = 0 looks_4 None of t...