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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. |
4. Write the first five terms of the G.P. where nth termis given as:521(i) 4.39-12n] |
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Answer» (1) nth term of GPTn = 4.3^n-1 Thus,T1 = 4.3^1-1 = 4*1 = 4T2 = 4.3^2-1 = 4*3 = 12T3 = 4.3^3-1 = 4*9 = 36T4 = 4.3^4-1 = 4*27 = 108T5 = 4.3^5-1 = 4*81 = 324 (2) nth term of GPTn = 5^n-1/2^n+1 Thus,T1 = 5^0/2^2 = 1/4T2 = 5^1/2^3 = 5/8T3 = 5^2/2^4 = 25/16T4 = 5^3/2^5 = 125/32T5 = 5^4/2^6 = 625/64 tysm |
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| 2. |
find the radius of sphere whose surface area is 154 cm |
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| 3. |
If $\cos \theta=-\frac{1}{\sqrt{2}},$ then $\cdot \theta^{\prime}$ is equal to :\begin{array}{lll}{\text { (A) } \frac{\pi}{3}} & {\text { (B) }-\frac{\pi}{3}} & {\text { (C) } \frac{2 \pi}{3}} & {\text { (D) } \frac{\pi}{6}}\end{array} |
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| 4. |
The area of a circle is 154 cm. Find its diameter. |
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| 5. |
The value of \left| \begin{array}{lll}{a-b} & {b-c} & {c-a} \\ {x-y} & {y-z} & {z-x} \\ {p-q} & {q-r} & {r-p}\end{array}\right| is equal to |
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| 6. |
\left[ \begin{array}{lll}{2} & {1} & {3}\end{array}\right]+\left[ \begin{array}{lll}{0} & {0} & {0}\end{array}\right] |
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Answer» same entires are equated [ 2 1 3] + [ 0 0 0] [ 2 1 3] |
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| 7. |
)If the 1st term of an A.P is 2, then show that to-I is five times the common difference.10 5 |
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Answer» if a=2t10=a+9dt5=a+4dt10-t5=a+9d-a-4d=5d i.e. 5 times the common difference |
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| 8. |
SECTION : B11.If the 10thterm of an A.P. is 47 and its 1st term is 2, find the sum of its first 15 terms. |
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Answer» First term a = 2let common difference = dTn = a + (n-1)d47 = 2 + (10-1)dd = 5Sn = n[2a + (n-1)d]/2S15 = 15[2*2 + 14*5]/2S15 = 15*37 = 555 |
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| 9. |
(f(x)= x 34 621 t are, in [1,3) Roue'sTheorem any = 241/3 a= ? any 6= ? |
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| 10. |
Between 1 and 31, m.numbers have been inserted in such a way that the resultingsequence is an A. P. and the ratio of 7'h and (m-1)h numbers is 5:9. Find thevalue of m.and (m - |
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| 11. |
LILLLLLLL LLLLL LLL LLLLLLLLL2. In a two-digit number, the digit at the units place is double the digit in the tens place. The numberexceeds the sum of its digits by 18. Find the number. |
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| 12. |
5 6 71. IfA =\left[ \begin{array}{lll}{2} & {4} & {6} \\ {3} & {5} & {7}\end{array}\right], B=\left[ \begin{array}{lll}{5} & {6} & {7} \\ {7} & {8} & {9}\end{array}\right], \text { then find } A+B |
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| 13. |
an A.P, the first term is 2 and the sum of the first five terms is one-fourth ofthe next five terms. Show that 20th term is-12 |
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| 14. |
22. If the first term of an A.P. is 2 and the sum of first five termsis equal to one-fourth of the sum of the next five terms, findthe sum of first 30 terms. |
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| 15. |
n an A.P the first term is 2 and the sum of the first five terms is one fourth of the next five termsShow that 20th term is -112 |
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| 16. |
umAP,thefirsttermis2and the sum of the first five terms is one-fourth ofmminIthe next five terms. Show that 20h term is -112. |
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| 17. |
ofIn an A.P, the first term is 2 and the sum of the first five terms is one-fourththe next five terms. Show that 20h term is-112.3, |
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| 18. |
1. Write the first five termsnth terms are- 2n-31) 2, |
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Answer» First five terms= n= 1,2,3,4,5so2-3/6= -1/6second term4-3/6= 1/6third term= 6-3/6= 3/6Fourth term= 8-3/6= 5/6fifth= 10-3/6= 7/6 |
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| 19. |
7. . In an A.P whose first term is 2, the sum of the first five terms is one fourth the sumof the next five terms. Show that a.112 Also find S |
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| 20. |
20. A number is 56 greater than the average of its third,quarter and one-twelfth. Find it. |
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| 21. |
A number is 56 greater than the average of its third, quarter and one-twelfth part. Find the number |
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Answer» Let the unknown number be 'x'Let the sum of terms be 'S'Let the no.of terms be 'n'Let the average be 'A'Average=Sum of terms/No.of termsSum of terms=One third of x+Quarter of x+One twelfth of xS=(x/3)+(x/4)+(x/12)S=(4x+3x+x)/12S=8x/12S=2x/3n=3A=(2x/3)×(1/3)A=2x/9As per the problem, the unknown number is 56 more than averagei.e x=56+(2x/9)=> x-(2x/9)=56=> (9x-2x)/9=56=> 7x/9=56=> 7x=504=> x=72Therefore the required number is 72 |
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| 22. |
e eighth term of an A.P. is half of its second term and the eleventh term exceeds oneird of its fourth term by 1. Find the 15th term.Find the arithmetic progression whose third term is 16 and seventh term exceeds itsNCERT]CBSE 2004]fifth term by 12.e ca rind the A.P |
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Answer» a + 7d = a + d/22a + 14d = a + da = -13da +10d = 1/3(a +3d) + 1a +10d -1 = 1/3(a + 3d)3(a + 10d -1) = a +3d3a + 30d -3 = a+ 3d2a + 27d =32(-13d) + 27d =3(since a = -13d)-26d + 27d = 3d = 3a= -13da = -39a + 14d = -39 + 42hence a15 =3 |
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| 23. |
24. Find the middle term(s) of the sequence 7, 13, 19.,... 241. |
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Answer» Given : An= 241a=7d = a2-a1 = 13-7= 6 An = a+(n-1)d 241= 7+(n-1)6241-7 = (n-1)6234/6 = n-139 = n-139+1 = 40 = n here n = number of terms An = last term a= first termd = difference with the 2 terms Total number of terms = 40Middle term = 40/2 = 20th An= 7 + (20-1)6 = 7 + (19)6 = 7 + 114 = 121 thus 121 is the middle term. |
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| 24. |
30. If the 1st term of a series is 7 and 13*term is 35. Find the sum of 13 terms ofICBSE 2012the sequence |
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Answer» Explanation : 1st term is 7 13th term is 35 Sum of 13 terms = 13/2 * (1st term + 13th term) = 13/2 * (7+35) = 13/2 * 42 = 13*21 = 273 Answer : 273 If you find this answer helpful then like it. |
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| 25. |
30. If the 1st term of a series is 7 and 13thterm is 35. Find the sum of 13 terms of[CBSE.2012]the sequence. |
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Answer» nth term of AP = a + (n-1)d 13th term = a + 12d Given a = 7 35 = 7 + 12d 12d = 28d = 28/12 = 7/3 Sum of n terms of AP = n/2(2a + (n-1)d) Sum of first 13 terms = 13/2(2*7 + 12*(7/3))= 13/2(14 + 4*7)= 13/2(14 + 28)= 13/2*(42)= 13*21= 273 ans |
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| 26. |
Ex. 9. Which fraction comes next in the sequence1 3 5 7?(M.A.T. 2005)2'4'8'16 |
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Answer» a1 = 1/2a2 = 1/2 + 1/4 = 3/4a3 = 3/4 + 1/8 = 5/8a4 = 5/8 + 1/16 = 7/16a5 = 7/16 + 1/32 = 9/32 Therefore next term in sequence is 9/32 Numerator = n + 2 [ where n = -1,1,3,5 7 ]Denominator = 2 * k [ where k = 1,2,4,8,16 ]So,Next term = 7 / 2 * 1 / 16 + 2 / 2 * 1 / 16 = 9 / 32. 9/32 is the answer.numerator= sequence of odd numbers.denominator= number ×2 9/32is the correct answer...... |
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| 27. |
2. The 20th term of the sequence 7, 3,-1,-5, ...is |
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Answer» a = 7 d = 3 - 7 = -4 tn = a + (n-1)d t20 = a + 19d = 7 + 19*(-4) = 7 - 76 = -69 If you find this answer helpful then like it. 7 is write ans -69. is write |
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| 28. |
ind the mode and mehat UTThe runs scored in a cricket match by 11 players is as follows:6 15 120, 50, 100, S0, 10, 15, 8, 10, 15 |
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| 29. |
20. A producer has 30 and 17 units of labour and capital respectively which he can useto produce two types of goods X and Y. To produce one unit of X, 2 units oflabour and 3 units of capital are required. Similarly, 3 units of labour and 1 unit ofcapital is required to produce one unit of Y. If X and Y are priced at $100 and120 per unit respectively, how should the producer use his resources to maximisethe total revenue? Formulate the above problem as a LPP and solve it graphically.16) |
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| 30. |
√5+√10= |
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| 31. |
√5-√3= |
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Answer» √2 is the correct answer I don't have any answer |
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| 32. |
√240+√60 |
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| 33. |
7 The angle in radians between the hanof a clock at 7:20 P.M. is(A) radians (B radians(C) -radians (D)Ī radians·IS-7tTm |
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Answer» B) 100 degrees = 100 x (π/180) 100 degrees = 5 x (π/9) 100 degrees = (5/9)π |
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| 34. |
(â48p*) = (â9p) |
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Answer» (-48p⁴)÷(-9p²)16p²/3 |
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| 35. |
Q17. Angle of elevation of top of a tower from the foot of a hill is 30° and that oftop the hill from the foot of the tower is 450. If distance between their feet is 45m,then determine the height of the hill and the tower |
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| 36. |
ple 4 Find the equation of the circle which passes through the points Cr2xam3,4) and whose centre lies on the line x +y 2.3,4) and whose centre lies on th |
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| 37. |
Let ÎABC ~ ÎDEF, if ar (AABC)-100 cm2, ar (ADEF)196cm2 and DE-7,then find AB. |
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Answer» ABC ~ DEF AB/ABC = DE/DEF AB/100 = 7/196 AB = 100×7/196 = 100/28 = 25/7 cm |
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| 38. |
ar (ADEFQ.9, lif the zeroes of the polynomial x2+px + g are double in value to the zeroes of 22-5x-3find the value of p and q |
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Answer» For equation 2x^2 - 5x - 3 = 02x^2 - 6x + x - 3 = 02x(x - 3) + 1(x - 3)=0(2x + 1)(x - 3) = 0x = - 1/2, 3 Then zeroes of x^2 + px + q = 0are - 1 and 6Then, -p/1 = - 1 + 6 p = - 5q/1 = - 1*6 = - 6q = - 6 |
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| 39. |
Find the differential equation of all circles which passes through theorigin and whose centre lies or |
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| 40. |
3 \sqrt [ 3 ] { 40 } - 4 \sqrt [ 3 ] { 320 } |
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| 41. |
3 \cdot \sqrt [ 3 ] { 40 } - 4 \cdot \sqrt [ 3 ] { 320 } - \sqrt [ 3 ] { 5 } |
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| 42. |
५६ ३ ३ १९९५ १ . ) ५४॥ ११(७ ।। ११ ।३ ५५ ६३ ।।५।।६३ ५ १३५ ५५' ब-१२।। ५।।-१३ 1 ॥६।। '५ ।०१)COSBC= 3 UT ACES |
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Answer» please post a clear image.... BC=3,CA=5this is right angled triangleAC square =AB square+BC square by Pythagoras theorm5 square=AB square+3 square25=ABsquare+9AB square=25-9=√16AB=4 |
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| 43. |
. Find the equation of the circle which touches the axes and whose centre lies onx-2y 3 |
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| 44. |
Find the differential equation of all circles which passes through theorigin and whose centre lies on y-axis. |
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| 45. |
12. Find the equation of the circle with radius 5 whose centre lies on x-axis andpasses through the point (2,3). |
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| 46. |
25 19-712, के लिए ज्ञात कीजि1. आव्यूह 4= 35 -21/31 -5 17](i) आव्यूह की कोटि(i) अवयवों की संख्या(ii) अवयव 41, 4, 43 44 43 |
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| 47. |
In ABC right-angled at B, AB = 24 cm, BC = 7cm. Determine(i) sin A. cos A (i) sin C. cos Cut frahanmand B- |
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| 48. |
A man depnumber of daysfor which the principal was depositedswhich is given to the student every year What is the rate of interest?of 4500 is deposited in a bank for the Best Student Award in a school. This amount earns annterest of ?540, |
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Answer» use this formula Rate of interest = simple interest × 100 ÷principle × time(assuming p = amount ) |
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| 49. |
if 5/6 gallon of paint covers 1/4 of the house, then how much paint is needed for the entire house? |
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Answer» 5/6 gallon paint requires for 1/4 housePaint needed for entire house = 4x5/6= 10/3= 3.33 gallons |
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| 50. |
ind the value of θ 172 sin 2θ = 13. |
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