#1
11th April 2016, 04:21 PM
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Simplification Question for SSC CGL
Hi I would like to have some questions on Simplification topic for the Staff Selection Commission’s Combined Group Level Examination?
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#2
11th April 2016, 04:21 PM
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Re: Simplification Question for SSC CGL
Some questions on Simplification topic for the Staff Selection Commission’s Combined Group Level Examination are as fillows: QUERY 1 If 2x + 2/x = 1, then find the value of x^3 + 1/x^3 Answer: whenever such type of question is asked …..first make it simple by converting the equation like: x+1/x = 1/2 Cubing it on both sides: (x + 1/x)^3 = (1/2)^3 = x^3 + (1/x)^3 + 3*x*1/x(x+1/x) = (1/2)^3 = x^3 + 1/x^3 = 1/8 – 3*1/2 Hence the expression to be find = -11/8 QUERY 2 If a^2 = 2, then find the value of a+1 a – 1 2/(a -1) (a + 1)/(3 – 2a) (a – 1)/(3 – 2a) Answer: Just put a+1 equal to all the options one by one till the result is a^2 = 2 1) a+1 = a – 1 impossible 2) a+1 = 2/(a – 1) a^2 – 1 = 2; a^2 = 3 (no solution) 3) a + 1 = (a + 1)/3 – 2a a = 3/2 (no solution) We see first 3 solutions are not there; so last one (option ‘4’) is the desired option… To me it seems the fastest method in such a question QUERY 3 If 1.5x = 0.04y, then the value of (y^2 – x^2) / (y^2 + 2xy + x^2) 730/77 73/77 73/770 74/77 Answer: 1.5 X = 0.04 Y therefore x/y = 2/75 The given expression = (y+x)(y-x)/(y+x)^2 = (y – x)/(y + x) = (1 – x/y)/(1 + x/y) put the value of x/y=2/75 = (1 – 2/75)/(1 + 2/75) = 73/75*75/77 = 73/77 (option ‘2’) Ans QUERY 4 If x^2 + y^2 – 2x + 6y + 10 = 0, then the value of x^2 + y^2 is Answer: x^2 + y^2 – 2x + 6y + 10 = 0 >> (9x^2 – 2x + 1) + (y^2 + 6y + 9) = 0 >> (x – 1)^2 + (y + 3)^2 = 0 Here we see that both the term oh LHS are square of a number, hence both are positive. We also know that if sum of the two positive numbers is zero both are zero. Therefore here both (x – 1)^2 = 0 and (y + 3)^2 = 0 >> x=1 , y=-3 Hence x^2 + y^2 = 10 QUERY 5 If x^4 + 1/x^4 = 119; then x^3 – 1/x^3 = ? Answer: x4 + 1/x4 = 119 (x2 + 1/x2)2 – 2 = 119 (x2 + 1/x2)2 = 121 x2 + 1/x2 = 11 (x – 1/x)2 + 2 = 11 (x – 1/x)2 = 9 x – 1/x = 3 Now to be valued: x3 – 1/x3 = ? (x – 1/x)3 + 3.x.1/x(x – 1/x) 33 + 3.3 27 + 9 36 QUERY 6 If X : Y=5 : 6, then (3x^2 – 2y^2) : (y^2 – X^2) is? 11:3 3:11 6:7 Answer: From the ratio of X and Y if X = 5 then Y = 6 Now putting these value of X and Y in in the other expression of ratios [3 (5)^2 – 2 (6)^2] : (6^2 – 5^2) ==> (75 – 72) : (36 – 25) ==> 3 : 11 (option ‘3) QUERY 7 If a : b : c = 2 : 3 : 4 and 2a – 3b + 4c = 33, then the value of c is 6 9 12 66/7 Answer: Put a, b, c as 2k, 3k, 4k in the given equation Then 11k = 33 ==> k = 3 ==> But ‘c’ = 4k ==> ‘c’ = 4 x 3 = 12 (option ‘3’) QUERY 8 If x + y + z = 1, xy + yz + zx = -1, xyz = -1, then what is the value of x^3 + y^3 + z^3 = ? 0 -1 -2 1 Answer: (x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx) => x^2 + y^2 + z^2 = 3 we know, x^3 + y^3 + z^3 – 3xyz = (x+y+z) (x^2 + y^2 + z^2 – xy -yz -zx) Now just put all the given & derived values, we get: x^3 + y^3 + z^3 = 1 (option ‘4’) TRICK By looking at the answer option we can easily derive that values of x, y, & z must be in integers. Now xyz = -1 (given) ; means none of them can be ‘0’; and either ‘all three are -1’ or ‘only one of them is -1 and two are 1 each’. If we see x^3 + y^3 + z^3 it’s not possible that x, y, z all are -1.. So two of them is 1 each and the third is -1. Therefore answer 1 is possible only. QUERY 9 If x^(1/3) + y^(1/3) = z^(1/3); then (x + y – z)^3 + 27xyz is equal to? Answer: The given equation = [x^(1/3) + y^(1/3)]^3 = z ==> x + y + 3x^2/3 y^1/3 + 3x^1/3 y^2/3 = z ==> x + y + 3x^(1/3) y^(1/3) [x^(1/3) + y^(1/3)] = z ==> x+y+ 3x^(1/3) y^(1/3) z^(1/3) = z [putting x^(1/3) + y^(1/3) = z^(1/3)] ==> x + y – z = -3 x^1/3 y ^1/3 z ^1/3 ==> (x + y – z )^3 = -27 xyz (cubing both sides) ==> (x + y – z)^3 + 27xyz = 0. QUERY 10 x^2 + y^2 + 1/x^2 + 1/y^2 = 4 then the value of x^2 + y^2 is? Answer: The given equation is = (x^2 + 1/x^2) + (y^2 + 1/y^2) = 4 ==> [(x – 1/x)^2 + 2] + [(y – 1/y)^2 + 2] = 4 ==> (x – 1/x)^2 + (y – 1/y)^2 + 4 = 4 ==> (x – 1/x)^2 + (y – 1/y)^2 = 0 We see that both the left hand terms are positive and their sum is equal to zero So (x – 1/x) = 0 and (y – 1/y) = 0 Solving both x^2 = 1 and y^2 = 1 Hence x^2 + y^2 = 2 |
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