Q: Question 2Dev brought an electrical fan which has a marked price of ₹ 800. If the GST on the goods is 7%, then the SGST is :₹ 24₹ 28₹ 56₹ 80

Since, GST rate is 7%, then SGST = 72\dfrac{7}{2}27 = 3.5%.M.P. of electrical fan = ₹ 800.SGST = 3.5% of ₹ 800= 3.5100×800\dfrac{3.5}{100} \times 8001003.5×800= ₹ 28.Hence, Option 2 is the correct option.

Q: Question 8If -4x > 8y; thenx > 2yx > -2yx < -2yx < 2y

Solving,⇒ -4x > 8y⇒ 4x < -8y⇒ x < −8y4-\dfrac{8y}{4}−48y⇒ x < -2y.Hence, Option 3 is the correct option.

Q: Question 9The value/s of 'k' for which the quadratic equation 2x2 - kx + k = 0 has equal roots is (are) :0 only4, 08 only0, 8

Given,Equation : 2x2 - kx + k = 0a = 2, b = -k and c = k.For equal roots,⇒ Discriminant (D) = 0⇒ b2 - 4ac = 0⇒ (-k)2 - 4 × 2 × k = 0⇒ k2 - 8k = 0⇒ k(k - 8) = 0⇒ k = 0 or k - 8 = 0⇒ k = 0 or k = 8.Hence, Option 4 is the correct option.

Q: Question 10If x = -2 is one of the solutions of the quadratic equation x2 + 3a - x = 0, then the value of 'a' is :-8-2−13-\dfrac{1}{3}−3113\dfrac{1}{3}31

Given,x = -2 is one of the solutions of the quadratic equation x2 + 3a - x = 0.∴ (-2)2 + 3a - (-2) = 0⇒ 4 + 3a + 2 = 0⇒ 3a + 6 = 0⇒ 3a = -6⇒ a = −63-\dfrac{6}{3}−36 = -2.Hence, Option 2 is the correct option.

Question 1

Question 1A retailer buys an article at its listed price from a wholesaler and sells it to a consumer in the same state after marking up the price by 20%. The list price of the article is ₹ 2500, and the rate of GST is 12%. What is the tax liability of the retailer to the central government?₹ 0₹ 15₹ 30₹ 60

Accepted Answer

For retailer,C.P. = ₹ 2500G.S.T. = 12%C.G.S.T. = 122\dfrac{12}{2}212​ % = 6%.C.G.S.T. paid = 6100×2500\dfrac{6}{100} 	imes 25001006​×2500 = ₹ 150.Given,Retailer sells the article to the consumer in the same state after marking up the price by 20%.S.P. = ₹ 2500 + 20%= ₹ 2500 + 20100×2500\dfrac{20}{100} 	imes 250010020​×2500= ₹ 2500 + ₹ 500= ₹ 3000.C.G.S.T. = 6%C.G.S.T. charged = 6100×3000\dfrac{6}{100} 	imes 30001006​×3000 = ₹ 180.Tax liability = C.G.S.T. charged - C.G.S.T. given= ₹ 180 - ₹ 150= ₹ 30.Hence, Option 3 is the correct option.

Question 2

Question 2Dev brought an electrical fan which has a marked price of ₹ 800. If the GST on the goods is 7%, then the SGST is :₹ 24₹ 28₹ 56₹ 80

Accepted Answer

Since, GST rate is 7%, then SGST = 72\dfrac{7}{2}27​ = 3.5%.M.P. of electrical fan = ₹ 800.SGST = 3.5% of ₹ 800= 3.5100×800\dfrac{3.5}{100} 	imes 8001003.5​×800= ₹ 28.Hence, Option 2 is the correct option.

Question 3

Question 3₹ P is deposited for n number of months in a recurring deposit account which pays interest at the rate of r% per annum. The nature and time of interest calculated is :compound interest for n number of monthssimple interest for n number of monthscompound interest for one monthsimple interest for one month

Accepted Answer

In a recurring deposit account,By formula,Interest = P×n(n+1)2×12×r100P 	imes \dfrac{n(n + 1)}{2 	imes 12} 	imes \dfrac{r}{100}P×2×12n(n+1)​×100r​∴ The nature and time of interest calculated is simple interest for one month.Hence, Option 4 is the correct option.

Question 4

Question 4Anwesha intended to open a Recurring Deposit account of ₹ 1000 per month for 1 year in a Bank, paying a 5% per annum rate of simple interest. The bank reduced the rate to 4% per annum. How much must Anwesha deposit monthly for 1 year so that her interest remains the same?₹ 12325₹ 1250₹ 1200₹ 1000

Accepted Answer

In first case :P = ₹ 1000r = 5%n = 12 monthsInterest = P×n(n+1)2×12×r100P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}P×2×12n(n+1)​×100r​=1000×12×(12+1)2×12×5100=1000×12×132×12×120=25×13=₹325.= 1000 \times \dfrac{12 \times (12 + 1)}{2 \times 12} \times \dfrac{5}{100} \[1em] = 1000 \times \dfrac{12 \times 13}{2 \times 12} \times \dfrac{1}{20} \[1em] = 25 \times 13 \[1em] = ₹ 325.=1000×2×1212×(12+1)​×1005​=1000×2×1212×13​×201​=25×13=₹325.In second case :P = ₹ x (Let)r = 4%n = 12 monthsInterest = ₹ 325∴325=x×12×132×12×4100⇒325=x×132×125⇒x=325×25×213⇒x=1625013=₹ 1250.\therefore 325 = x \times \dfrac{12 \times 13}{2 \times 12} \times \dfrac{4}{100} \[1em] \Rightarrow 325 = x \times \dfrac{13}{2} \times \dfrac{1}{25}\[1em] \Rightarrow x = \dfrac{325 \times 25 \times 2}{13} \[1em] \Rightarrow x = \dfrac{16250}{13} = \text{₹ 1250.}∴325=x×2×1212×13​×1004​⇒325=x×213​×251​⇒x=13325×25×2​⇒x=1316250​=₹ 1250.Hence, Option 2 is the correct option.

Question 5

Question 5Mr. Das invests in ₹ 100, 12% shares of Company A available at ₹ 60 each. Mr. Singh invests in ₹ 50, 16% shares of Company B available at ₹ 40 each. Use this information to state which of the following statements is true.The rate of return for Mr. Das is 12%The rate of return for Mr. Singh is 10%Both Mr. Das and Mr. Singh have the same rate of return of 10%Both Mr. Das and Mr. Singh have the same rate of return of 20%

Accepted Answer

For Mr. Das,Face value of each share = ₹ 100Market value of each share = ₹ 60Dividend per share = 12% of ₹ 100 = ₹ 12.Rate of return = Dividend per shareM.V. of each share×100=1260×100\dfrac{	ext{Dividend per share}}{	ext{M.V. of each share}} 	imes 100 = \dfrac{12}{60} 	imes 100M.V. of each shareDividend per share​×100=6012​×100 = 20%.For Mr. Singh,Face value of each share = ₹ 50Market value of each share = ₹ 40Dividend per share = 16% of ₹ 50 = ₹ 8.Rate of return = Dividend per shareM.V. of each share×100=840×100\dfrac{	ext{Dividend per share}}{	ext{M.V. of each share}} 	imes 100 = \dfrac{8}{40} 	imes 100M.V. of each shareDividend per share​×100=408​×100 = 20%.∴ Both Mr. Das and Mr. Singh have the same rate of return of 20%.Hence, Option 4 is the correct option.

Question 6

Question 6Amit invested a certain sum of money in ₹ 100 shares, paying a 7.5% dividend. The rate of return on his investment is 10%. The money invested by Amit to purchase 10 shares is :₹ 250₹ 750₹ 900₹ 1100

Accepted Answer

Let ₹ P be the price per share.Amit brought 10 shares, so total investment = ₹ 10PDividend = 7.5%Dividend per share = 7.5100×100\dfrac{7.5}{100} \times 1001007.5​×100 = ₹ 7.5Total dividend = ₹ 7.5 × 10 = ₹ 75.Rate of return = 10%By formula,Rate of return × Investment = Dividend⇒10⇒10100×10P=75⇒100P100=75⇒P=₹75.\Rightarrow 10% \times 10P = 75 \[1em] \Rightarrow \dfrac{10}{100} \times 10P = 75 \[1em] \Rightarrow \dfrac{100P}{100} = 75 \[1em] \Rightarrow P = ₹ 75.⇒10⇒10010​×10P=75⇒100100P​=75⇒P=₹75.Total investment = 10P = 10 × ₹75 = ₹750.Hence, Option 2 is the correct option.

Question 7

Question 7If -3 ≤ -4x + 5 and x ∈ W, then the solution set is :{.......-3, -2, -1, 0, 1, 2, 3, .......}{1, 2}{0, 1, 2}{2, 3, 4, 5}

Accepted Answer

Solving,⇒ -3 ≤ -4x + 5⇒ 4x ≤ 5 + 3⇒ 4x ≤ 8⇒ x ≤ 84\dfrac{8}{4}48​⇒ x ≤ 2.Since, x ≤ 2 and x ∈ W.∴ Solution set = {0, 1, 2}.Hence, Option 3 is the correct option.

Question 8

Question 8If -4x > 8y; thenx > 2yx > -2yx < -2yx < 2y

Accepted Answer

Solving,⇒ -4x > 8y⇒ 4x < -8y⇒ x < −8y4-\dfrac{8y}{4}−48y​⇒ x < -2y.Hence, Option 3 is the correct option.

Question 9

Question 9The value/s of 'k' for which the quadratic equation 2x2 - kx + k = 0 has equal roots is (are) :0 only4, 08 only0, 8

Accepted Answer

Given,Equation : 2x2 - kx + k = 0a = 2, b = -k and c = k.For equal roots,⇒ Discriminant (D) = 0⇒ b2 - 4ac = 0⇒ (-k)2 - 4 × 2 × k = 0⇒ k2 - 8k = 0⇒ k(k - 8) = 0⇒ k = 0 or k - 8 = 0⇒ k = 0 or k = 8.Hence, Option 4 is the correct option.

Question 10

Question 10If x = -2 is one of the solutions of the quadratic equation x2 + 3a - x = 0, then the value of 'a' is :-8-2−13-\dfrac{1}{3}−31​13\dfrac{1}{3}31​

Accepted Answer

Given,x = -2 is one of the solutions of the quadratic equation x2 + 3a - x = 0.∴ (-2)2 + 3a - (-2) = 0⇒ 4 + 3a + 2 = 0⇒ 3a + 6 = 0⇒ 3a = -6⇒ a = −63-\dfrac{6}{3}−36​ = -2.Hence, Option 2 is the correct option.

Question 11

Question 11In solving a quadratic equation, one of the values of the variables x is 233.356. The solution rounded to two significant figures is :233.36233.35233.3230

Accepted Answer

x = 233.356On rounding off to two significant figuresx = 230.Hence, Option 4 is the correct option.

Question 12

Question 12In the adjoining diagram, AB = x cm, BC = y cm and x - y = 7 cm. Area of △ ABC = 30 cm2. The length of AC is :10 cm12 cm13 cm15 cm

Accepted Answer

By formula,Area of triangle = 12× base×height\dfrac{1}{2} 	imes 	ext{ base} 	imes 	ext{height}21​× base×height⇒ 30 = 12×BC×AB\dfrac{1}{2} 	imes BC 	imes AB21​×BC×AB⇒ 30 = 12×y×x\dfrac{1}{2} 	imes y 	imes x21​×y×x⇒ 30 = xy2\dfrac{xy}{2}2xy​⇒ xy = 60 ........(1)Given,⇒ x - y = 7⇒ x = 7 + y ........(2)Substituting value of x from equation (2) in (1), we get :⇒ (7 + y)y = 60⇒ 7y + y2 = 60⇒ y2 + 7y - 60 = 0⇒ y2 + 12y - 5y - 60 = 0⇒ y(y + 12) - 5(y + 12) = 0⇒ (y - 5)(y + 12) = 0⇒ y - 5 = 0 or y + 12 = 0⇒ y = 5 or y = -12.Since, side cannot be negative,∴ y = 5 cm.Substituting value of y in equation (2), we get :⇒ x = 7 + y = 7 + 5 = 12 cm.In right angle triangle ABC,By pythagoras theorem,⇒ AC2 = AB2 + BC2⇒ AC2 = x2 + y2⇒ AC2 = 122 + 52⇒ AC2 = 144 + 25⇒ AC2 = 169⇒ AC = 169\sqrt{169}169​ = 13 cm.Hence, Option 3 is the correct option.

Question 13

Question 13If p, q, and r are in continued proportion, then :p : q = p : rq : r = p2 : q2p : q2 = r : p2p : r = p2 : q2

Accepted Answer

Since, p, q, and r are in continued proportion.∴pq=qr⇒q2=pr ........(1)\therefore \dfrac{p}{q} = \dfrac{q}{r} \[1em] \Rightarrow q^2 = pr \text{ ........(1)}∴qp​=rq​⇒q2=pr ........(1)Solving,p : r = p2 : q2⇒pr=p2q2⇒q2=p2×rp⇒q2=pr .........(2)\Rightarrow \dfrac{p}{r} = \dfrac{p^2}{q^2} \[1em] \Rightarrow q^2 = \dfrac{p^2 \times r}{p} \[1em] \Rightarrow q^2 = pr \text{ .........(2)}⇒rp​=q2p2​⇒q2=pp2×r​⇒q2=pr .........(2)Since, equation (1) and (2) are equal.Hence, Option 4 is the correct option.

Question 14

Question 14The ratio of diameter to height of a Borosil cylindrical glass is 3 : 5. If the actual diameter of the glass is 6 cm, then the curved surface area of the glass is :120π60π30π18π

Accepted Answer

Given,Ratio of diameter to height of a Borosil cylindrical glass is 3 : 5.⇒dh=35⇒6h=35⇒h=6×53⇒h=10 cm.\Rightarrow \dfrac{d}{h} = \dfrac{3}{5} \[1em] \Rightarrow \dfrac{6}{h} = \dfrac{3}{5} \[1em] \Rightarrow h = \dfrac{6 \times 5}{3} \[1em] \Rightarrow h = 10 \text{ cm}.⇒hd​=53​⇒h6​=53​⇒h=36×5​⇒h=10 cm.Radius = Diameter2=62\dfrac{\text{Diameter}}{2} = \dfrac{6}{2}2Diameter​=26​ = 3 cm.By formula,Curved surface area of glass = 2πrh= 2π × 3 × 10= 60π cm2.Hence, Option 2 is the correct option.

Question 15

Question 15If the polynomial 2x3 + 3x2 - 2x - 3 is completely divisible by (2x + a), and the quotient is equal to (x2 - 1), then one of the values of a is :-3-113

Accepted Answer

Given,The polynomial 2x3 + 3x2 - 2x - 3 is completely divisible by (2x + a) and quotient is equal to (x2 - 1).∴ 2x3 + 3x2 - 2x - 3 = (2x + a)(x2 - 1)⇒ 2x3 + 3x2 - 2x - 3 = 2x3 - 2x + ax2 - a⇒ 2x3 - 2x3 + 3x2 - 2x + 2x - 3 = ax2 - a⇒ 3x2 - 3 = ax2 - aFrom above equation,a = 3.Hence, Option 4 is the correct option.

Question 16

Question 16A polynomial in x is x3 + 5x2 - kx - 24. Which of the following is a factor of the given polynomial so that the value of k is 2?(x + 2)(x - 3)(x + 4)(x - 4)

Accepted Answer

By factor theorem,(x - a) is a factor of f(x), if f(a) = 0.Given,Polynomial = x3 + 5x2 - kx - 24If k = 2, then :Polynomial = x3 + 5x2 - 2x - 24.Dividing polynomial by (x + 4) or substituting -4 in polynomial, we get :⇒ (-4)3 + 5(-4)2 - 2(-4) - 24⇒ -64 + 5(16) + 8 - 24⇒ -64 + 80 + 8 - 24⇒ 88 - 88⇒ 0.Since, on substituting -4 in polynomial, we get remainder = 0.∴ (x + 4) is the factor of x3 + 5x2 - kx - 24, when k = 2.Hence, Option 3 is the correct option.

Question 17

Question 17If A = [ab] and [cd]\begin{bmatrix*}[r] a & b \end{bmatrix*} 	ext{ and } \begin{bmatrix*}[r] c \ d \end{bmatrix*}[a​b​] and [cd​], then :only matrix AB is possibleonly matrix BA is possibleboth matrices AB and BA are possibleboth matrices AB and BA are possible, AB = BA

Accepted Answer

Order of matrix A = 1 × 2Order of matrix B = 2 × 1Since, no. of columns in A is equal to the no. of rows in B and no. of columns in B is equal to the no. of rows in A.∴ AB and BA are possible.AB=[a×c+b×d]=[ac+bd].BA=[c×ac×bd×ad×b]=[cacbdadb].AB = \begin{bmatrix*}[r] a \times c + b \times d \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] ac + bd \end{bmatrix*}. \[1em] BA = \begin{bmatrix*}[r] c \times a & c \times b \ d \times a & d \times b \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] ca & cb \ da & db \end{bmatrix*}.AB=[a×c+b×d​]=[ac+bd​].BA=[c×ad×a​c×bd×b​]=[cada​cbdb​].∴ AB ≠ BA.Hence, Option 3 is the correct option.

Question 18

Question 18Matrix A = [69−4k] such that A2=[0000]\begin{bmatrix*}[r] 6 & 9 \ -4 & k \end{bmatrix*} 	ext{ such that } A^2 = \begin{bmatrix*}[r] 0 & 0 \ 0 & 0 \end{bmatrix*}[6−4​9k​] such that A2=[00​00​]. Then k is :6-636±6

Accepted Answer

Given, A2=[0000]A^2 = \begin{bmatrix*}[r] 0 & 0 \ 0 & 0 \end{bmatrix*}A2=[00​00​].∴[69−4k][69−4k]=[0000]⇒[6×6+9×(−4)6×9+9×k−4×6+k×(−4)(−4)×9+k×k]=[0000]⇒[36+(−36)54+9k−24−4k−36+k2]=[0000]⇒[054+9k−24−4k−36+k2]=[0000]⇒54+9k=0⇒9k=−54⇒k=−549=−6.\therefore \begin{bmatrix*}[r] 6 & 9 \ -4 & k \end{bmatrix*}\begin{bmatrix*}[r] 6 & 9 \ -4 & k \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \ 0 & 0 \end{bmatrix*} \[1em] \Rightarrow \begin{bmatrix*}[r] 6 \times 6 + 9 \times (-4) & 6 \times 9 + 9 \times k \ -4 \times 6 + k \times (-4) & (-4) \times 9 + k \times k \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \ 0 & 0 \end{bmatrix*} \[1em] \Rightarrow \begin{bmatrix*}[r] 36 + (-36) & 54 + 9k \ -24 - 4k & -36 + k^2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \ 0 & 0 \end{bmatrix*} \[1em] \Rightarrow \begin{bmatrix*}[r] 0 & 54 + 9k \ -24 - 4k & -36 + k^2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \ 0 & 0 \end{bmatrix*} \[1em] \Rightarrow 54 + 9k = 0 \[1em] \Rightarrow 9k = -54 \[1em] \Rightarrow k = -\dfrac{54}{9} = -6.∴[6−4​9k​][6−4​9k​]=[00​00​]⇒[6×6+9×(−4)−4×6+k×(−4)​6×9+9×k(−4)×9+k×k​]=[00​00​]⇒[36+(−36)−24−4k​54+9k−36+k2​]=[00​00​]⇒[0−24−4k​54+9k−36+k2​]=[00​00​]⇒54+9k=0⇒9k=−54⇒k=−954​=−6.Hence, Option 2 is the correct option.

Question 19

Question 19If the sum of n terms of an arithmetic progression Sn = n2 - n, then the third term of the series is :2469

Accepted Answer

Given,Sum of n terms of an arithmetic progression Sn = n2 - n.S1 = 12 - 1 = 0,S2 = 22 - 2 = 4 - 2 = 2,S3 = 32 - 3 = 9 - 3 = 6.Sum upto first term = First term = 0.Given, sum upto 2 terms = 2 and first term = 0, second term = 2.Sum upto third term = 6∴ First term + Second term + Third term = 6⇒ 0 + 2 + Third term = 6⇒ Third term = 6 - 2 = 4.Hence, Option 2 is the correct option.

Question 20

Question 20Which of the following is not a geometric progression ?13,1,3,9\dfrac{1}{3}, 1, 3, 931​,1,3,915,15,15,15\dfrac{1}{5}, \dfrac{1}{5}, \dfrac{1}{5}, \dfrac{1}{5}51​,51​,51​,51​-2, 4, -8, 162, 0, 4, 0, 8, 0

Accepted Answer

In the series,2, 0, 4, 0, 8, 002≠40\dfrac{0}{2} 
e \dfrac{4}{0}20​=04​∴ 2, 0, 4, 0, 8, 0 is not a G.P.Hence, Option 4 is the correct option.

Multiple-Choice Questions