SSC CGL ALGEBRA MCQ ENGLISH

Correct Answer : (D)

Explanation : If each term of a given AP is multiplied or divided by a non-zero constant k, then the resulting sequence is also an AP with common difference k × d or kd , where d is the common difference of the given AP.If a, b and c are in AP then k – a, k – b, k – c are in AP where k ≠ 0.

Correct Answer : (B)

Explanation : We know that,If an object covers Dkm at a speed of S1 km/hr and next Dkm at a speed of S2 km/hr . Then the average speed of the object is given by: =2×S1×S2S1+S2It is given that,As we know that, if an object covers Dkm at a speed of S1 km/hr and next Dkm at a speed of S2 km/hr . Then the average speed of the object is given by: =2×S1×S2S1+S2Here, D=60 km, S1=3vkm/hr,S2=2vkm/hrThe average speed of the car =2×S1×S2S1+S2=2×3v×2v5v=2.4v km/hr

Correct Answer : (A)

Explanation : The general term in the expansion (1−x2)16 is given byTr+1=(−1)r16Cr(1)16−r(x2)r=(−1)r16Cr(x2r) …..(i)Now, we need the coefficient of x8On equating power of x in (i) with 8, we get2r = 8⇒ r = 4∴T4+1=(−1)416C4x8Coefficient of x8=16C4=16×15×14×134×3×2×1= 4 × 5 × 7 × 13 ….(ii)Now, in (x+2x)8, general term is given by,Tr+1=8Cr(x)8−r(2x)r=8Cr(x)8−rx−r2r=8Cr(x)8−2r2rNow, for an independent term power of x should be 0.So, 8 – 2r = 0⇒ 2r = 8⇒ r = 4∴T4+1=8C4(x)8−4(2x)4Coefficient of independent term =8C4×24=8×7×6×54×3×2×1×16= 2 × 7 × 5 × 16 …. (iii)So, Ratio =4×5×7×132×7×5×16=138

Correct Answer : (A)

Explanation : Given: A = {1, 2, 3, …., 14} and R is a relation defined on A such that R = {(x, y) : 3x – y = 0 where x, y ∈ A}When x = 1 ∈ A then y = 3x = 3 ∈ A ⇒ (1, 3) ∈ RWhen x = 2 ∈ A then y = 3x = 6 ∈ A ⇒ (2, 6) ∈ RWhen x = 3 ∈ A then y = 3x = 9 ∈ A ⇒ (3, 9) ∈ RWhen x = 4 ∈ A then y = 3x = 12 ∈ A ⇒ (4, 12) ∈ RSo, R = {(1, 3), (2, 6), (3, 9), (4, 12)}As we know that Range (R) = {b: (a, b) ∈ R}⇒ Range(R) = {3, 6, 9, 12}

Correct Answer : (B)

Explanation : It is given that,The constant term in the expansion of (x−kx2)10 is 405i.e the independent term of (x−kx2)10 is 405 .Let (r+1)th term be the independent term.⇒Tr+1=10Cr×(x)10−r×(−kx2)r=10Cr×(−k)r×(x)10−5r2∵(r+1) th term is the independent term⇒10−5r2=0⇒r=2⇒10C2×(−k)2=405⇒45k2=405⇒k=±3

Correct Answer : (C)

Explanation : It is given that,4(x – p)(x – q) – r2 = 0⇒ x2 – (p + q) x + pq – r24 = 0By comparing the quadratic equation x2−(p+q)x+(pq−(r2)2)=0 with the standard quadratic equation ax2 + bx + c = 0. We get, a = 1, b = – (p + q) and c = pq – r24⇒ D = b2 – 4ac= (p + q)2 – 4pq + r2= (p – q)2 + r2 ≥ 0Therefore, statement 1 is true.The given quadratic equation 4(x – p)(x – q) – r2 = 0 will have repeated roots if D = 0.⇒ D = (p – q)2 + r2 = 0 if and only if p = q and r = 0.Therefore, statement 2 is also true.

Correct Answer : (C)

Explanation : Let A = Students who play cricket, B = Students who play football⇒ n (A) = 50% and n (B) = 40%⇒ n (A ∩ B) = 10%⇒ n (A ∪ B) = n (A) + n (B) – n (A ∩ B) = 50 + 40 – 10 = 80%As we know that (A ∪ B) ’ = A’ ∩ B’⇒ n ( (A ∪ B) ’) =n (U) – n (A ∪ B) = 100 – 80 = 20% = n (A’ ∩ B’)

Correct Answer : (B)

Explanation : Given, |3x – 5| ≤ 2⇒ – 2 ≤ 3x – 5 ≤ 2⇒ – 2 + 5 ≤ 3x ≤ 2 + 5⇒ 3 ≤ 3x ≤ 7⇒1≤x≤73So, if |3x−5|≤2 then then 1≤x≤73

Correct Answer : (D)

Explanation : Here,(cdccddcccddd)2, where c > d⇒ (cdccddcccddd)2 = (101100111000)2⇒ (101100111000)2 = (1 × 211) + (1 × 29) + (1 × 28) + (1 × 25) + (1 × 24) + (1 × 23) + 0 = (2872)10

Correct Answer : (A)

Explanation : We know that,If f(x)=|x| thenf(x)={x,x≥0−x,x<0It is given that,x2 + 3|x| + 2 = 0Case- 1: when x ≥ 0⇒ x2 + 3x + 2 = 0⇒ x2 + x + 2x + 2 = 0⇒ (x + 1) (x + 2) = 0⇒ x = - 1 or - 2∵ x ≥ 0The given quadratic equation has no real roots for x ≥ 0.Case- 2: when x < 0⇒ x2 - 3x + 2 = 0⇒ x2 - x - 2x + 2 = 0⇒ (x - 1) (x - 2) = 0⇒ x = 1 or 2∵ x < 0The given quadratic equation has no real roots for x < 0.

Correct Answer : (B)

Explanation : We know that,If α and β are the roots of the quadratic equation, ax2 + bx + c = 0. Thenα+β=−ba and α×β=caIt is given that,p and q are the roots of the equation x2 – 30x + 221By comparing the given equation with the standard quadratic equation ax2 + bx + c = 0, we get a = 1, b = – 30 and c = 221.As we know that, if α and β are the roots of the quadratic equation, ax2 + bx + c = 0. Thenα+β=−ba and α×β=ca⇒p+q=30 and pq=221⇒p3+q3=(p+q)×(p2−pq+q2)=(p+q)×[(p+q)2−3pq]⇒p3+q3=(p+q)×[(p+q)2−3pq]=30×[900−663]=7110

Correct Answer : (B)

Explanation : Here, R = {(a, b): a,b ϵ Z and (a + b) is even}1. Since, a + a = 2a, ,which is even, so R is reflexive.2. If a + b is even then b + a will also be even. So, R is symmetric.3. Let, a = 3, b = 5, and c = 7.a + b = 3 + 5 = 8 (which is even), b + c = 5 + 7 = 12 (which is again even) and a + c = 3 + 7 = 10 (which is also even)Therefore, R is an equivalence relation on Z.

Correct Answer : (C)

Explanation : It is given that,xlog7⁡x>7By taking log both the sides we get,⇒(log7⁡x)×log⁡x>log⁡7As we know that, loga⁡b=log⁡blog⁡a⇒(log⁡x)2log⁡7>log⁡7⇒(log⁡x)2>(log⁡7)2⇒(log⁡x)2−(log⁡7)2>0⇒(log⁡x+log⁡7)×(log⁡x−log⁡7)>0x∈(0,17)∪(7,∞)

Correct Answer : (B)

Explanation : Let z = x + iyAccording to the question,arg⁡(z)=tan−1⁡yx=π3⇒yx=tan⁡π3⇒yx=3⇒y=3xAlso given xy=33⇒y=33xComparing the values of y, we get⇒33x=3x⇒x2=3⇒x=±3∴x=3{∵ arg (z) lies in first quadrant so -ve value is neglected}Also y=3x=3Complex number z=x+iy=3+3iConjugate of z=(z¯)=3−3i

Correct Answer : (D)

Explanation : Here, one root is 5p – qSince q is not a perfect square, therefore other root will be 5p + qNow, required quadratic equation is x2 – (sum of roots)x + product of roots = 0⇒ x2 – (5p + q + 5p – q )x + (5p + q )(5p – q ) = 0⇒ x2 -10px + 25p2 – q = 0⇒ x2 -10px + 25p2 – q = 0

Correct Answer : (A)

Explanation : Given: binomial expansion: (1+x)32Tr+1=32Cr(1)n−r(x)r=32×(32−1)×…(32−r+1)r!×(x)rNow, for this term to become negative,(32−r+1)<0⇒r>52⇒ r > 2.5⇒ r = 3 (∵ r is never fractional)⇒ r + 1 = 3 + 1 = 4T3+1=32×(32−1)×(32−2)3!×(x)3=32×12×(−12)3×2×(x)3=−x316

Correct Answer : (B)

Explanation : We know that,Mean of N observations is given by: ∑i=1nxiNIt is given that mean weight of 150 students in a certain class is 60 kg .Let M be the mean weight of 150 students⇒M=∑i=1150xi150=60⇒∑i=1150xi=9000 —–(1)Let there are x no. of boys ⇒ The no .of girls = 150 – xIt is given that mean weight of the boys is 70 kg and that of the girls is 55 kgLet M2 = 70 and M3 = 55.The total weight of all the boys = 70xThe total weight of all the girls = 55 × (150 – x) = 8250 – 55xThe total weight of all the students = 8250 – 55x + 70x = 8250 + 15x ———-(2)Using equation (1) and (2), we get⇒ 8250 + 15x = 9000 ⇒ x = 50.∴ No. of boys = 50 and No. of girls = 150 – 50 = 100.

Correct Answer : (C)

Explanation : We know that,nPr=n!(n−r)! and nCr=n!r!×(n−r)!nPrnCr=r!It is given that,P (n, r) = 2520 and C (n, r) = 21As we know that, nPrnCr=r!⇒P(n,r)C(n,r)=252021=120=r!⇒r=5As we know that, nPr=n!(n−r)!⇒nP5=n!(n−5)!=2520⇒n=7⇒C(n+1,r+1)=C(8,6)⇒8C6=8!6!×(8−6)!=28

Correct Answer : (C)

Explanation : The given equation is,C(47, 4) + C(51, 3) + C(50, 3) + C(49, 3) + C(48, 3) + C(47, 3)= {C(47, 4) + C(47, 3)} + C(51, 3) + C(50, 3) + C(49, 3) + C(48, 3)As we know that, C (n, r – 1) + C (n, r) = C (n + 1, r)⇒ {C(47, 4) + C(47, 3)} + C(51, 3) + C(50, 3) + C(49, 3) + C(48, 3)= {C (48, 4) + C(48, 3)} + C(51, 3) + C(50, 3) + C(49, 3)⇒ {C (48, 4) + C(48, 3)} + C(51, 3) + C(50, 3) + C(49, 3) = { C(49, 4) + C(49, 3)} + C(51, 3) + C(50, 3)⇒ {C(49, 4) + C(49, 3)} + C(51, 3) + C(50, 3) = {C(50, 4) + C(50, 3)} + C(51, 3)⇒ {C(50, 4) + C(50, 3)} + C(51, 3) = { C(51, 4) + C(51, 3)} = C(52, 4

Correct Answer : (A)

Explanation : We know that,If a is the first term and r is the common ratio of a GP then the nth term of GP is given by:an = arn – 1It is given that,3rd, 8th and 13th terms of a GP are p, q and r respectively.Let a is the first term and R is the common ratio.⇒ a3 = aR2 = p —(1)⇒ a8 = aR7 = q —-(2)⇒ a13 = aR12 = r —(3)Dividing (2) by (1), we get⇒ qp = R5 —(4)Similarly, by dividing (3) by (2), we get⇒rq=R5 —(5)From (4) and (5), we get⇒ q2 = p × r

Correct Answer : (D)

Explanation : It is given that,S={2,4,6,8,………..20}⇒ The no. of elements in set S,n(S)=10As we know that, for any non-empty set A with n(A)=x . Then the total no. of subsets of set A is given by:2x⇒ The maximum number of subsets of S=210=1024

Correct Answer : (C)

Explanation : The general term of (x−kx2)9 is given by,Tr+1=(−1)r×9Cr×(x)9−r×(kx2)r=(−1)r×9Cr×(x)9−r×x−2r×kr=(−1)r×9Cr×(x)9−r×x−2r×kr=(−1)r×9Cr×(x)9−3r×krNow, the term is free from x (independent term), then power of x is zero.On equating power of x from (1) with zero, we get9 – 3r = 0⇒ r = 3⇒ r + 1 = 4∴T3+1=(−1)3×9C3×(x)9−9×k3=−9×8×73×2×1×k3= -84k3∴ -84 = -84k3⇒ k = 1

Correct Answer : (D)

Explanation : 1. We know, a×b=ab only when a,b≥0−a×−b=−1a×−1b=ab×(i×i)….(∵i=−1)=−ab….(∵i2=−1)2. i4m+3i4m+3 = i4mi3= i3 ….(∵ i4m = 1)= -i ….(∵ i3 = -i)So, both the statements are not correct.

Correct Answer : (C)

Explanation : It is given that, m and n (m < n) be the roots of the equation x2 - 16x + 39 = 0⇒ x2 - 13x - 3x + 39 = 0⇒ x (x - 13) - 3 (x - 13) = (x - 13) × (x - 3) = 0⇒ x = 13 or 3⇒ m = 3 and n = 13∵ m, p, q, r, s and n are in AP⇒ p + q + r + s = (m + d) + (m + 2d) + (m + 3d) + (m + 4d)= 4m + 10d= 2 × (2m + 5d)= 2 × [m + (m + 5d)]= 2 × (m + n) ∵ (n = m + 5d)⇒ p + q + r + s = 2 × (m + n) = 32 ∵ m + n = 16

Correct Answer : (B)

Explanation : Two digit numbers which are divisible by 4 are: 12, 16, 20, …, 96 forms an AP with first term a = 12, common difference d = 4 and nth term an = 96.⇒ an = a + (n – 1) × d = 96⇒ 12 + (n – 1) × 4 = 96⇒ n = 22

Correct Answer : (A)

Explanation : Let N denote the set of natural numbers⇒ N = {1, 2, 3, 4,….}A = {n2 : n ∈ N}⇒ A = {1, 4, 9, 16, ….}B = {n3 : n ∈ N}.⇒ B = {1, 8, 27, 64, ….}Consider the statement “A ∪ B = N”⇒A ∪ B = {1, 4, 8, 9, 16, 27,….} ≠ NSo, the statement A ∪ B = N is not true.Consider the statement “The complement of (A ∪ B) is an infinite set”We know that, The complement of a set, denoted A’, is the set of all elements in the given universal set U that are not in A.⇒ (A ∪ B)’ = U – (A ∪ B) = {2, 3, 5, 6,….} = infinite set⇒ (A ∪ B)’ = Infinite setTherefore, the statement “The complement of (A ∪ B) is an infinite set” is true.Consider, the statement “A ∩ B must be a finite set”A ∩ B = {1}So, the statement “A ∩ B must be a finite set” is true.Consider, the statement “A ∩ B must be a proper subset of {m6 : m ∈ N}”Let S = {m6 : m ∈ N}S = {1, 64, ….}and A ∩ B = {1}A ∩ B must be a proper subset of {m6 : m ∈ N}Therefore, the statement “A ∩ B must be a proper subset of {m6 : m ∈ N}” is True.So, if N denote the set of natural numbers and A = {n2 : n ∈ N} and B = {n3 : n ∈ N}. then the statement (2), (3) and (4) are correct.

Correct Answer : (B)

Explanation : A∪B={x:(x−a)⋅(x−b)>0;aa has two signs in range (a,∞)∴ x>a is not possible.(b) xb has only one sign in range (b,∞)∴ x>b also satisfies.∴A={x:xb}