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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
f a, b are positive real numbers, a > b and a2+ b27ab, prove thatlog a + log b) |
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| 2. |
laws of exponents applied to negative exponents |
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Answer» Negative Exponent Rule: This says that negative exponents in the numerator get moved to the denominator and become positive exponents. Negative exponents in the denominator get moved to the numerator and become positive exponents. Only move the negative exponents. A negative exponent means to divide by that number of factors instead of multiplying. So 4^−3 is the same as 1/(4^3), and x^-3 = 1/x^3. As you know, you can’t divide by zero. So there’s a restriction that x^−n = 1/x^n only when x is not zero. When x = 0, x^−n is undefined. |
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| 3. |
exponents means |
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Answer» The exponent of a number says how many times to use that number in a multiplication. It is written as a small number to the right and above the base number. |
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| 4. |
All the laws of exponents |
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Answer» aⁿxaᵐ=aⁿ+ᵐaⁿ/aᵐ=aⁿ-ᵐaⁿxbⁿ=(axb)ⁿ(aⁿ)ᵐ=aⁿᵐ Laws of Exponents Bases – multiplying the like ones – add the exponents and keep base same. (Multiplication Law) Bases – raise it with power to another – multiply the exponents and keep base same. Bases – dividing the like ones – ‘Numerator Exponent – Denominator Exponent’ and keep base same. (Division Law) Let ‘a’ is any number and ‘m’ , ‘n’ are positive integers, then Multiplication Law am× an = am+n Division Law am÷an = am/ an = am-n Negative exponent a-m= 1/am Exponent Rules i) a0= 1 ii) (am)n= a(mn) iii) am× bm=(ab)m iv) am/bm= (a/b)m Laws of Exponents- (i)am/an= am-n (ii)a(m)n= amn (iii)am× bm= (ab)m (iv)am/bm= (a/b)m (v)a0= 1 (vi) m√(bn) =bn/m (vii) b1/n=n√b (viii) b-n= 1 /bn Laws of Exponents. When multiplying like bases, keep the base the same and add theexponents. When raising a base with a power to another power, keep the base the same and multiply theexponents. When dividing like bases, keep the base the same and subtract the denominatorexponentfrom the numeratorexponent. Accordingto the product law of exponents when multiplying two numbers that have the same basethen we can add the exponents am× an=am+n where a, m and n all are natural numbers. Here the base shouldbe the same in both the quantities. For example, 2³× 24= 27 22/3× 21/5= 22/3 + 1/5 = 2(10+3)/15. We get, = 212/15 (-6)3x (-6)2= (-6)3+2= (-6)5 |
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| 5. |
Arun purchased 60 kg vegetable at Rs.420,then sold them at the rate at Rs.6.50 perk.g. Then the profit percent is |
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| 6. |
4500 and sold aof the goods at a loss of 10%At what gain percent should heAman bought goods worthsell the remaining goods so as to gain 20% on the whole transaction?12. |
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| 7. |
a.Salrmcb.Differentc. Lessthall30. Abhay bought a scooter for Rs 45000 and after a years ssold it at a loss of 15% find the S.P31,225a. Rs 32850b. Rs 38250c. Rs 31,225d. |
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| 8. |
TUSP01 ()11. A's income is 10% more than B's. How much per centis B's income is less than A's?(a) 10%(b) 7% |
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Answer» A's income is 10 % more than B . So B's income less than A is. = 10 / ( 10 + 100 ) × 100 %= 10/110 × 100 %= 1000/110 %= 9.09 % |
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| 9. |
TU60. A does 20% less work than B. If A cancomplete a piece of work in 7 hrs, then B can doit in |
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Answer» Ratio of percentage of work done by A and B = 80:100Suppose B takes 'x' hours time to complete the work.So,⇒ 80:100::x:15/2⇒ 100x = 15/2*80⇒ x = 600/100⇒ x = 6 hours (Time taken by B) |
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| 10. |
PHASE-I FUN BOOK4 MARKS:9. Graph:MATHEMATICS CBSEFor the following distribution, draw a ‘ less than type, ogive and from the curve, find the median.Marks Obtained Less Less Less Less Less Less ess Less Lessthan than than than hanthan tha than20 30405060 7080 90 100No. of Students 2 7 17 40 6 82 85 90 100 |
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| 11. |
निम्नलिखित समीकरण निकायों को वज़है. x + 8y =192x + 11y = 28 |
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| 12. |
)Form two linear equations to find incomes from the following information:A and B have incomes in the ratio 3:4 and their expenses are in the ratio 2:3, and eachsaves 8000 |
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Answer» Let incomes of A and B are 3x and 4x Let expenses of A and B are 2y and 3y Then,3x - 2y = 8000.....(1)4x - 3y = 8000.....(2) Multiply eq(1) by 3 and eq(2) by 29x - 6y = 24000....(3)8x - 6y = 16000....(4) Subtract eq(4) from eq(3), we getx = 24000 - 16000 = 8000 Put value of x in eq(1)3*8000 - 2y = 80002y = 16000y = 8000 Income of A = 3*8000 = 24000Expense of A = 2*8000 = 16000 Income of B = 4*8000 = 32000Expense of B = 3*8000 = 24000 |
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| 13. |
It becomesnumeratan and 2 is subtichfractiontwo pessons is an and thefind the2fdach of themratio of incomes of27 Thetheir expenditure is 4:3ofannual incomesdatiotheiryearly findsaves Rs 2000thrice as old as B. andCA wasshall be twice as old as B. Whatin van hence Aof A and Be |
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Answer» what is your question |
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| 14. |
Rewrite each of the expressions below in a simpler form. Your newexpressions should not contain negative exponents.a. (5x3X-3x2)b. (4p2q)-1 |
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| 15. |
Q. 2. How does noise pollution affect our i |
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Answer» Ans :- Noise isany disturbing or unwantedsound, andnoise pollution affectspeople's health and quality oflife. Prolonged highlevelsofnoise cancause hearing loss and stress-related illnesses.Noiseoftenaffects children more than adults, andnoise pollutionalsoaffectsgeneral well-being. एल्कीन श्रेणी का सामान्य सूत्र लिखिए |
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| 16. |
values of sinQ10, in Δ。PQ, right-angled at P, OP-7cm and OQ-PQ = 1cm. Determine the valueand cosQProve that : sin':9 + cos"θ + 3sinag cosag!11y) |
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Answer» Consider, sin6ө+cos6ө =1-3sin2өcos2ө(sin2ө)3+(cos2ө)3= (sin2ө+cos2ө)3− 3 (sin2ө cos2ө)(sin2ө+cos2ө) [since a + b = (a+b)3− 3ab(a+b)] = 1 − 3sin2өcos2ө [Since sin2ө+cos2ө = 1] OPQ is rt angled at P .OP=7cmOQ-PQ=1cmso, OQ = 1 + PQBY PYTHAGORUS THEOREM,OQ^2=OP^2+PQ^2(1+PQ)^2= 7^2 + PQ^21 + PQ^2+2PQ= 49 + PQ^22PQ= 48PQ=24cmOQ= 25cm thanks |
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| 17. |
ery shot Anser lype Questions1. Find the number nearest to 110000 but greater than 100000 which is exactlydivisible by each of 8, 15 and 21. |
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Answer» 109200 |
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| 18. |
Long Answer lype Questions23, Rahul is constructing his house. He wants to make a hall of dimensions 20 m x 15(5 marks each)m, such that it hasborder of black granite 0.5 m wide on all four sides and a black granite circle in the middle of radius1.4 m. Calculate the cost of granite used at1200 per square metre |
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| 19. |
50 shares at 8%and the other part is invested in 8% 25 shares at 8% discount, then theDivide20304 into two parts such that if one part is invested in 900annual incomes from both the investments are equal |
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Answer» Let the two parts in which Rs.20304 is to be divided be Rs. x and Rs.(20304 - x). In First Case, Investment = Rs.x.Rate of Dividend = 9%.Nominal Value of 1 Share(N.V.) = Rs. 50Market Value of 1 Share(M.V.) = 50 + 50 × 8/100= 50 + 4= Rs. 52∵ Number of Shares = Sum Invested/M.V. of 1 Share∴ No. of Shares = x/54 ∵ Annual Income = No. of Shares× Rate of Dividend× N.V. of 1 Share.∴ Annual Income = x/54× 9/100× 50 ⇒ Annual Income = x/12 In Second Case, Investment = Rs. (20304 - x)N.V. of 1 Share = Rs. 25M.V. of 1 Share = 25 - 25× 8/100= 25 - 2= Rs. 23Rate of Dividend = 8 %No. of Shares = (20304- x)/23 Annual Income = (20304 - x)/23× 8/100× 25= Rs. 2(20304 - x)/23 According to the Question,Annual Income from both Investments are equals.∴ x/12 = 2(20304 - x)/23⇒ 23x = 24(20304) - 24x⇒ 24x + 23x = 24× 20304⇒ 47x = 487296⇒ x = Rs. 10368 Thus, First part is Rs. 10368. Second Part = 20304 - 10368= Rs. 9936 |
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| 20. |
1) A & B invested RS 2000 & RS 5000. So if there is a profit of RS 28000what is A's share if they share the profits in the ratio 2:5? |
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Answer» they share profit in ratio = 2:5then profit of "A" = 2/5 so profit of "A" =28000×2/7 = 8000 ₹ they share profit in ratio =2.5then profit of 'A' = 2/5 so profit of 'A' = 28000×2/7 = 8000Rs ratio = 2000 : 5000 = 2 : 5 A's share = ( A's Ratio / (A + B Ratio ) ) * Profit = ( 2 / (2+5) ) * 28000 = ( 2 / 7 ) * 28000 = ( 2 ) * 4000 = 8000 So the answer is Option ( c ) 8000Also if you need B's = 28000 - 8000 = 20000 20000 rupya correct answer for this question 20000 rupaiye correct answer hoga like my comment 20,000 is the right answer. |
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| 21. |
o 18. Sunil invested Rs. 2000 for 3 years and Rs. 3000 for 22 years at the same rainterest. He received a total interest of Rs. 810, what was the rate of interest? |
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Answer» SI = P*R*T/100 For P= 2000, T = 3, R=? SI1 = 2000*R*3/100 = 60R For P = 3000, T = 5/2, R =?SI2 = 3000*R*5/200 = 75R According to the given conditionSI1 + SI2 = 81060R + 75R = 810135R = 810R = 810/135 = 6 Rate of Interest = 6% |
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| 22. |
2) Smita has invested Rs.12,000 and purchased shars of FV Rs. 10 at a premium ofRs.2. Find number of shares she purchased. Complete th activitySoln: FV- Rs.10Premium Rs.2Total investmentMVNumber of shares12000Ans. Smita has purchased ⥠shares. |
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Answer» MV = FV + premium = 10 + 2 = 12 Number of shares = Total Investment /MV= 12000/12= 1000 Smita has purchased 1000 shares Like my answer if you find it useful! |
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| 23. |
38) Mr. Thomas Invested an amount of Rs. 13,900divided in two different schemes A and B atthe simple interest rate of 14% p.a. and 11%p.a, respectively. If the total amount of simpleInterest earned in 2 years be Rs. 3508, whatwas the amount invested in Scheme B?a. Rs. 6300b. Rs. 6400c. Rs. 6500d. Rs. 6600hatun simple and20) TheBee |
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| 24. |
बुक, व न ध्न् > \67 : ’2.- gt : N V’p L?WA TR ™ ey पृ] |
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Answer» Like if you find it useful |
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| 25. |
8,00,0003,20,000On 1st April, 2018, they decided to sell a machine for 2,00,000 which was purchased1st April, 2015 for 3,00,000. Prepare Machinery Account and Provision forDepreciation Account for the year ended 31st March, 2019 assuming that the firm hasbeen charging depreciation 10% p.a. on the Straight Line Method.Solution |
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| 26. |
से निम्नलिखित में प्रत्येक का विस्त(i) (5p—3g)" « |
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| 27. |
ILLUSTRATION 15.Rajan and Rajani are partners in a firm. Their capitals were-Rajan 3,00,000;Rajaniき2,00,000. During the year ended 31st March, 2016 the firm earned a profit of1,50,000. Calculate the value of goodwill of the firm on the basis of capitalisation ofaverage profits assuming that the normal rate of return is 20%. |
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| 28. |
atedonthe. On April 01, 2010, Bajrang Marbles purchased a Machine for Rs. 2.80.000and spent Rs. 10,000 on its carriage and Rs. 10,000 on its installation. It isestimated that its working life is 10 years and after 10 years its scrap valuewill be Rs. 20,000,(a)Prepare Machine account and Depreciation account for the st fouryears by providing depreciation on straight line method. Acc ints areclosed on March 31st every yeardepreciation account forare closed on March 31 every yearotalcitlypare Machine account, Depreciation account and Provision foraccumulated depreciation account) for the first(b) Prefour years by providing depreciation using straight line method accounts(Ans: laBalance of Machine account on April 1, 2014 Rs. 1.28.0o0bl Balance of Provision for depreciation account as on 1.04.2014Rs.72,000.) |
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| 29. |
4u square+8u |
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Answer» The equation is 4u2+8uTaking 4u common4u(u+2)=0U+2=0U=-2. Hope it helps |
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| 30. |
23. 4u +17-7x8x + 5y -19 |
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| 31. |
\begin{array} { l } { \text { EXAMPLE 4 Prove that: } } \\ { \text { (i) tan } ^ { - 1 } \left\{ \frac { \sqrt { 1 + \cos x } + \sqrt { 1 - \cos x } } { \sqrt { 1 + \cos x } - \sqrt { 1 - \cos x } } \right\} = \frac { \pi } { 4 } - \frac { x } { 2 } , \text { if } \pi < x < \frac { 3 \pi } { 2 } } \end{array}{ 2 } |
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| 32. |
\operatorname { cos } \{ 2 [ \operatorname { Tan } ^ { - 1 } \frac { 1 } { 4 } + \operatorname { tan } ^ { - 1 } \frac { 2 } { 9 } ] \} |
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| 33. |
\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right) |
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Answer» tan⁻¹1/4 + tan⁻¹ 2/9 = tan⁻¹( 1/4 + 2/9)/(1-(1/4)*(2/9))= tan⁻¹ (1/2) let tan⁻¹(1/2)=θ So, tanθ = 1/2So, we can find cos θ = 2/√5cos 2θ = 2cos²θ -1 = 2*4/5 - 1 = 3/5So, cos 2θ = 3/5hence, 2θ = cos ⁻¹ 3/5 θ = 1/2 cos⁻¹ 3/5but, θ = tan⁻¹1/2 So, tan⁻¹ 1/2 = 1/2 cos⁻¹ 3/5 |
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| 34. |
2. Find the perimeter of givenfigure, where AED is asemicircle and ABCD is arectangle. |
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| 35. |
\tan \left( \cos ^ { - 1 } \frac { 4 } { 5 } + \tan ^ { - 1 } \frac { 2 } { 3 } \right) |
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Answer» thanks |
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| 36. |
7. In the adjoining figure, if AB LBC and DE 1 AC. Prove thatA ABC-A AED.DVBO |
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| 37. |
23. Find the perimeter of a given fig, where AED is a semicircle and ABCD isrectangle.20 cm寸20 cm |
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| 38. |
MATHEMATICS-D2In Fig. 16.201, ABCD is a quadrilateral inscribed in a circle with centre 0. Crproduced to E such that <AED = 95° and <OBA 30°. Find OAC.30°Fig. 16.201 |
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| 39. |
Find the zeroes of the following quadratic polynomials seen the zeroes and the coefficients.(i)x^2 - 2x -8(ii)4s^2 - 4s + 1(iii)6x^2 - 3 - 7x(iv)4u^2 + 8u |
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| 40. |
21.ABC is a right triangle, right angled at B. If P is the length of the perpendicular from B toAC and AB =C , BC = a and CA = b , then prove that (a)==+= (b) p?b? = aục?nie te verse kleure is the length of the perpendicular con |
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Answer» 1.Area is 1/2 base * height . So for the same triangle area can be 1/2*b*a and can be 1/2*p*c. Hence cp = ab 2 . From above p2 = a2b2/c2 1/p2 = c2/ a2b2 but for right triangle c2 = a2 +b2 Hence 1/p2 = (a2+b2)/ a2b2 Hence 1/p2 = 1/a2+1/b^2re |
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| 41. |
VERSE TRIGONOMETRIC FUNCTIONS6. Prove that:1 2 1 3(i) tan 1 1tancos4 |
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| 42. |
What single transformation was applled to trlangle A to get trlangle BtChoose 1 answerA) Translation |
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Answer» complete question please dilation is the correct answer |
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| 43. |
13. What should be added to13 -348 1 to get 1 ? |
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Answer» thanks |
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| 44. |
At the end of 2years on ₹ 2000 at 5%p.a. compound annually |
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Answer» amount is=2205 and compound interest=205 i hope you understand yes Compound interest = Amount = 2000(1+5/100)^22205then interest will be2205-2000205 rupees |
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| 45. |
OTTUIUIU-JIE 6.The difference between the compound interest, compounded andsimple interest on a certain sum for 2 years at 15% per annumthe sum.180. Find2 years at 15% per annum is |
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Answer» Let ₹P be the principal. Then, S.I. = P×R×T/100 = P×15×2/100 = ₹3P/10 C.I. = P{(1+R/100)^n - 1} = P{(1+15/100)^2 - 1} = P{(100+15/100)^2 - 1 = P{(115/100)^2 - 1} = P{(23/20)^2 - 1} = P{529/400 - 1} = P{529 - 400/400} = P{129/400} = ₹129P/400 According to question, C.I. - S.I. = 180 => 129P/400 - 3P/10 = 180 => 129P - 120P/400 = 180 => 9P/400 = 180 => P = 180×400/9 => P = ₹8,000 Hence, the sum is ₹8,000. |
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| 46. |
Find the sum that yields an interest of 170 after 2yearsat the rate of 121% pa. interest being compoundedannually. |
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| 47. |
ShodiyaTaand Kegan want toratio 54 How muchdivide Rs. 90.000will each getने |
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Answer» 5x+4x=90,0009x=90,000x=90,000/9x=10,0004x=4*10,000=40,0005x=5*10,000=50,000 |
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| 48. |
Example 10, Mukul invests 9000 in a company paying a dividend of 6% per annumwhen a share of face value 100 stands at 150, What is his annual income? He sells 50% ofhis shareswhen the price rises to 200. What is his gain on this transaction? |
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Answer» 1 2 |
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| 49. |
(2) Mr. Mahajan purchased 100 shares, cach of face valuc Rs. 100the market price was Rs 45 per share, paying 2% brokerage. If the rateof GST on the brokerage is 18%, find the total amount he spent. |
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Answer» Like if you find it useful |
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| 50. |
man has 300, 50 shares of a company20% dividend. Find his net incomepayingafter paying 3% income tax |
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Answer» Total share price = 300*50 = 15000Dividend = (20/100)*15000 = 3000Income tax = (3/100)*3000 = 90Net income = 3000 - 90 = Rs 2910 |
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