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.
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Date2.2Page |
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Answer» x²+x+5=0 ->1 To find x we have to factorise equ 1 Discriminant = b²-4ac a=1;b=1;c=5. Discriminant = 1-(4×1×5) Discriminant = -19 x1= (-b + √discriminant)/2×a x1= (-1+19i)/2. x2= (-b - √discriminant)/2×a x2= (-1 - 19i)/2. Hencex1= (-1+19i)/2. and x2= (-1 - 19i)/2. |
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| 2. |
What is the length of a cuboid with width 13 cm, height 10 cm, and volume1690 cm? |
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Answer» 13x10=130:; 1690/130=13 Given,Width = 13 cmHeight = 10 cmVolume = 1690 cm³Now,find length:Length = Volume / Area=> 1690 / 130 = 13.°. Length = 13 cm |
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PageDate2 S농- |
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Answer» y = 25 ( vertically opposite angle)x+ 25 = 180 ( straight line) x = 180 - 25 = 155so z= 155 ( vertically opposite angle) thx sir |
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| 4. |
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Answer» (a² + b²) ( -a² + b²) (b² + a²) ( b² - a²) (b²)² - (a²)² b⁴ - a⁴ |
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| 5. |
PAGE NO:DATE:212 |
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Answer» 1 2 3 |
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| 6. |
The number of coins 1.5 cm in diameter and 0.2 cm thick to be meltedto form a right circular cylinder of height 10 cm and diameter 4.5 cm is' |
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| 7. |
ëld-u, 22_c.in e tehfinal |
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Answer» Thank u so much |
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| 8. |
find the number of coin of diameter 1.5 and 0.2 CM thick to be melted to form of right circular cylinder of height 10 cm and diameter 4.5 CM |
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| 9. |
. Solve the equation :(i) log10 (x + 1) + log10 (x - 1)= log 10 11 + 2 log103(CSE) |
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Answer» plzz answer anyone i will be helpfull too u plzzzz |
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| 10. |
4. Simplify :2 . e 36४208 .452 270 0 (8l)4_ 25)7 - (5)_M e — कि दी कलदर (1) 16X 2n+__4 X 8n (” 16 9 दी . 2 |
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| 11. |
1. If (0 2)* 2 and log10 2 03010, then what 4is the value of x to the nearest tenth?(a) -10.0(b) 0-5(c) -0-4(d)02 |
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Answer» (0.2)^x = 2 taking log to both the sides xlog(0.2) = log2 => xlog(2/10) = log2=> x[log2-log10] = log2=> x[0.30-1] = 0.30=> x = -0.30/0.70 => x = -3/7 = -0.42 = -0.4 option C |
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| 12. |
1. If (0-2) 2 and log10 2 03010,theis the value of x to the nearest tenth?(a) 10.0(b) -05(e)04(d) 02 |
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Answer» (0.2)^x= 2 taking log on both the sides xlog(0.2) = log2=> x[log(2/10)] = log(2)=> x[ log2-1] = log(2)=> x = log2/log2-1 = 0.30/0.30-1 = -0.30/0.70 ≈ -0.43 option c |
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| 13. |
C in Fig. 6.56, Ps is the bisector ofZQPR of△PQR. Prove that QS-mSR PR |
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Answer» Like if it is useful |
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| 14. |
4.Solve: 16x-15 =65 |
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Answer» 16x - 15 = 6516x = 65 + 1516x = 80 x = 80/16 x = 5 |
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| 15. |
(1) In the figure, seg Ps i side QR.If PQ- a, PR b, QS c and RS-dthen prove that(a + b) (a-b) = (c + d) (c-d). |
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46.Inthegivenfigure(not drawn to scale), ind the valueof (b + d) (a +c).50°al1150 |
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Answer» thanku |
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| 17. |
(I) In the figure, seg PS i side QR.if PQ a, PR b, QS c and RS- dthen prove that(a+b(a -b) (c+d)(c d). |
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| 18. |
4*u^2 %2B 8*u |
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| 19. |
-09-0zagil.sal.322. A और B साथ मिलकर एक काम को 40 दिनों में कर सकते हैं।उनकी काम करने की दर का अनुपात 8:5हैं। A उसी काम को• अकेला कितने दिनों में पूरा करेगा?IANCERPIAN INDIA |
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Answer» the correct answer is 104 days the correct answer is 104 days |
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| 20. |
\frac { 1 } { 4 } \cdot \frac { 1 } { 2 } , \frac { 3 } { 4 } , 1 , \frac { 5 } { 4 } \dots \ldots \text { is } \frac { 25 } { 4 } ? |
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Answer» a=1/4d=1/2-1/4=1/425/4=a+(n-1)d25/4=1/4+(n-1)1/46*4=n-1n-1=24n=25so 25th term |
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| 21. |
\frac{7}{8} \div \frac{5}{4}+\frac{3}{5} \times \frac{1}{4}-\frac{3}{10} |
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| 22. |
\frac { 2 } { 5 } \times \frac { 4 } { 9 } \div \frac { 4 } { 7 } + \frac { 3 } { 5 } - \frac { 2 } { 5 } |
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| 23. |
CHAPTER - SQUARE AND SQUARE ROOTSQ.1 If m is the cube root of n , then(b) vm(a) m3(d) Ým(c) |
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| 24. |
10. Find quadratic equation such that its roots are square of sum of the roots andsquare of difference of the roots of equation 2x^2 + 2(p + q) x +p^2+q^2 = 0 |
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| 25. |
४ 6370 - 4218 + 845 2 5की. डे _'..“. |
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| 26. |
विवार को 845 व्यक्ति चिड़ियाघर गए। सोमवार को केवल 169 व्यक्ति गए। चिडियाघरकी सैर करने वाले व्यक्तियों की संख्या में सोमवार को कितने प्रतिशत कमी हुई? |
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Answer» monday=845-169=676Monday=(676/845)*100=80% 60.76 loss is the correct answer me Monday=845-169=676 (676/845)*100 =80% 80% correct answer for this question 80% is correct answer Answer:Monday=845-169676 (676/845)*100=80% 80% the correct answer of the given question |
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| 27. |
(4)15x2-16x + 1-0 |
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Answer» 15x² - 16x +1=0=> 15x² -15x -x +1=0=> 15x ( X -1) -1(x-1)=0=> (15x -1)(x-1)=0=> X = 1/15, 1. Please hit the like button below if this helped you |
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| 28. |
foctorize the following.(1)16x^2+24yx+9y^2 |
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Answer» 16x²+24xy+9y²=(4x)²+(2×4×3)xy +(3y)² Thus it is of form a²a+2ab+b² Where a=4x; b=3y Thus a²+2ab+b²=(a+b)²now (4x)²+(2×4×3)xy +(3y)²=(4x+3y)² 16x^2+24xy+9y^2(4x)^2+2*4x*3y+(3y)^2(4x+3y)^2 |
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| 29. |
nuit be addad to4x4+x1So that theu 8 |
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Answer» p( x )= 4x⁴+ 2x³- 2x²+x -1 g(x )= x²+2x-3 x²+2x-3)4x⁴+2x³-2x²+x-1(4x²-6x+22 4x⁴+8x³-12x²______________________-6x³+10x²+x -6x³-12x²+18x___________________ 22x²-17x-1 ____________________-61x+65 The remainder = -61x +65 Now we have to add 61x -65 to p(x) So that the resulting polynomial is divisible by g( x ) thanks |
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| 30. |
\frac { 5 } { 4 } + ( \frac { - 11 } { 4 } ) |
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| 31. |
\frac 5 4 %2B ( \frac - 11 4 ) |
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Answer» 5/4 + (-11/4)=(5-11)/4= -6/4 = -3/2 answer is -6 upon 4 because base is same and plus and minus make minus so it is correct |
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| 32. |
roclice set 7.5Yield al Selabean pecMuind the meanamedSayabean 1s. 875 |
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Answer» Mean = Summation of yield / Summation of acre= (10+7+5+3+9+6+9+6)/8= 55/8= 6.875 Please hit the like button if this helped you |
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| 33. |
CHAPTER - SQUARE AND SQUARE ROOTSQ.1 If m is the cube root of n, then(a) m3 (b) Vm(b) vm(0)(d) Vm(a) m3 |
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Answer» answer: (a)coz 3√n. = mhence m^3 = n option d is correct for this question then n = m^3 correct ans option a is correct answer |
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| 34. |
€91 (W) LSO ()uशिव फिर 03% 5. |
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| 35. |
\frac{5}{4}+\left(\frac{-11}{4}\right) |
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Answer» 5/4 - 11/4 = -6/4 = -3/2 is the answer |
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| 36. |
शा LL ( (४)पड 2० ०0६ 14% ८0:70Dl e 1किफिरि धुंध पुर भर 128 k DBk T |
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| 37. |
4x4 +42 16x-4If x +4, then find the value of |
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| 38. |
Seltmi danesd A lso par mmh in a bank& smanths unde |
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Answer» THE total principle amount for 8 months is 150*8 = 1200Rs = P , R= 8% , T = 8 months = 8/12 years now interest gained = P*R*T/100 = 1200*0.08*2/3 = 64rs so, total maturity value is 1200+64 = 1264rs |
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C.B.S.E. 2015)2 in the adjoining figure, O is the centre of a circle of radius 5 cmT is a point such that OT = 13 cm and OT intersects circle at E.!AB is a tangent to the circle at E, find the length of AB, wheTP and TQ are two tangents to the circle(CBSE, 20 |
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| 40. |
find the value of log10 0.001 |
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Answer» c=-3 is the correct answer of the given question |
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| 41. |
an soĹcales,es, .tiangs AGC, AC AC.LA 80ăŤ. |
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| 42. |
Find the degree of each term for the following polynomial and also find the degree ofthe polynomial |
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Find the zeroes of the following polynomial:5V5x2+30x +85 |
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Find the zeroes of the following polynomial:5v5x2+30x+8v5 |
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3) Write the following polynomial in standarm3 +4 +3m. |
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Answer» In standard form the terms of the polynomial should be in decreasing order of their exponents.So, in standard form the following polynomial is:-m³+3m+4 |
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(-11)/4 %2B 5/4 |
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| 47. |
चŕĽ. 2B Dl *O9x=((z-) . |
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Answer» x² - 2x = (-2) ( 3 - x) x² - 2x = -6 + 2x x² - 4x + 6 = 0 So, given is has degree 2 so it is quadratic equation |
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| 48. |
x³ +4x²+ 5x +7 |
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| 49. |
Find the value of p for which the points (5, 1), (1, p) and (4, -2) arecollinear. |
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Answer» x₁ = -5 x₂ = 1x₃ = 4 y₁ = 1y₂ = py₃ = -2 For the points to be collinear :- x₁(y₂ -y₃) + x₂(y₃ - y₂) + x₃(y₁ - y₂) = 0 -5{ p -(-2)} + 1 ( - 2 - 1) + 4( 1 - p) = 0 -5(p+2) + 1(-3) + 4( 1 - p) = 0 -5p - 10 - 3 + 4 - 4p = 0 -9p - 9 = 0 -9p = 9 p = 9/-9 p = -1 |
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| 50. |
ik by (2 |k DL u l_E |
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Answer» π/3 =180/3 =60° |
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