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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. |
The printed price ofa book is? 1 50. And discount is l 5%. Find thepaid.The markedprie c5. |
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Answer» 0.15×150=22.5150-22.5=127.5 where is steps sir sir reply u want answer or steps steps hello sir please repu exam vundi hello steps evi Ra bai |
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| 2. |
3. In A ABC, AB = AC, ZA = 2x° and ZB = x + 10°. Find themeasures of all the angles of A ABC. |
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Answer» In triangleAB= AC then angle B = angle Cnow2x+2(x+10)= 1802x+2x+20= 1804x= 160x= 40°angle B= angle C= 50angle A= 80 |
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| 3. |
In a quadrilateral ABCD, ZA = 72ยบ and C is the supplementary of LA. The other twoangles are 2x-10 and x + 4. Find the value of x and the measure of all the angles. |
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Answer» I don't know answer so please tell me you A=72 B=114 C=108 D=66 X=62 angle c=180-72=108°now,2x-10+x+4+72+108=3603x-6=1803x=186x=62° A=72 B=114 C=108 D=66 X=62 angle A=72 degreeangle Cis the supplement of A=180-72=108angle C =108degreesum of all angles=360 degree72+108+2x-10+x+4=360180-10+4+2x+x=360166+3x=3603x=360-1663x=294x=294/3x=98angleB=2×98-10angle B=186degreeangle D=98+4angle D=102degree |
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| 4. |
In figure 11.21, one pair of adjacent sides ofa parallelogram is in the ratio 3:4. If one ofits angles, ZA is a right angle and diagonal 3xBD = 10 cm, find the-9.A l(i) lengths of the sides of the parallelogram.4x |
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Answer» thanks bro |
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| 5. |
(-1 %2B sqrt(2))/(1 %2B sqrt(2)) %2B (1 %2B sqrt(2))/(-1 %2B sqrt(2)) |
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Answer» (root(2) - 1)/(root + 1) +(root(2) +1)/(root - 1) Take LCM = (root(2) - 1)^2 + (root(2) + 1)^2/ (root(2)^2 - 1^2 ) = (2 + 1 + 2root(2)) + (2 + 1 - 2root(2))/(2 - 1) = (3 + 3) / 1 = 6 |
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| 6. |
५५५।।।Using completing the square method, show that the equationx4 - 8x + 18 = 0 has no solution. |
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Answer» To complete the square, we need the perfect square of the equation ofx^2 - 8x + 18In order to find the perfect square, we need to change the equation into(x−b)^2 = a, were a and b are constants. To find c, we divide the coefficient by 2 and square it (8/2)^2 = 16 We get 16, which means that we must change our current equation to have a 16. x^2 - 8x + 18 − 2 = −2 By subtracting 2 from both sides, we get that 16. Now, we can simplify the left hand side into the perfect square x^2 - 8x + 16 = (x-4)^2 This means(x-4)^2 = −2 We now square root both side, giving usx - 4 = √−2 They can never be a negative square root, so therefore there is no answer |
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| 7. |
4x^2+4 root3x+3=0do it using completing sq method |
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Answer» part 1 part 2 |
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| 8. |
Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Heip him by completing the following table. |
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Answer» thx |
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| 9. |
58. एक परीक्षा में, रमेश ने गुरदास से 15 अंक अधिक पाए, जबकिलीला ने उसी परीक्षा में गुरदास से 7 अंक कम पाए। यदि उनकेकुल प्राप्तांक 83 हैं, तो गुरदास द्वारा प्राप्तांक ज्ञात कीजिए।।(A) 18(B) 25(C)40(D)35 |
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Answer» option b is the right answer option b is the right answer option b is the right answer option b is the right answer 18 is the right answers 25 is the correct answer of this question option, b 25 is the right answer the correct answer is B 25 is the right answer. d is the right answer a is the correct answer option A is write answer 25 is the correct answer of the question 25 is tha correct answer of the question |
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| 10. |
5' ७ ८, 445. लुप्त संख्या ज्ञात कीजिए।128...16, 7.251हैं।|A2-c(a) 184__(b) 210~ 241) 425 |
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Answer» 425 is the correct answer the correct answer is 425 184 12² - 8² = 8016² - 7² = 20725² - 21² = 184 |
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| 11. |
o. AB is a line segment and P is its mid-point. Dand E are the points on the same side ofA such that < BAD = < ABE and < EPA =(a) Δ DAP EBP(b) ADBEDPB. Show that |
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| 12. |
AB is a line segment and P is its mid-point. D anE are points on the same side of AB such tha< BAD = < ABE and EPA = DPB(see Fig. 7.22). Show that(i) ΔDAP ΔΕΒΡ(ii)AD=BE |
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| 13. |
ERAMPLE 4. |
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| 14. |
7.AB is a line segment and P is its mid-point. D andE are points on the same side of AB such thatBAD-< ABE and < EPA = 2 DPB(see Fig. 7.22). Show that(i) ΔDAP ΔΕΒΡ(ii) AD-BE |
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| 15. |
AB is a line segment and P is its mid-poinE are points on the same side of AB such thatt. D and7,BAD = < ABE and < EPA = < DPB(see Fig. 7.22), Show that(i) Δ DAP Δ ΕΒΡ(ii) AD-"BEFig, 7.22 |
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| 16. |
Triangle ABC is right angled at A. AD is drawn perpendicular to BC. lf AB 5 cm andAC = 12 cm, find the area of ΔABC. Also find the length of AD |
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| 17. |
7. Verify thefollowing using the algebraic identitiesa) (a 7)-84a (3a 7)Q Find the continued nroduuet |
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| 18. |
25. Find the middle terms of the A.P. 7,13,19,. 241 |
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| 19. |
25.Find the middle terms of the A.P 7, 13, 19,,241 |
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Answer» a=7d=6 241=a+(n-1)d241=7+(n-1)6234/6=n-139=n-1n=40 Therefore,last term of the given AP is 40th term Middle term would be the 20th term 20th term=7+(20-1)6=7+19*6=7+114=121 |
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| 20. |
The identity used to find the roots of a quadratic eguation in one varable usingcompleting the square is(A) (a t b)2(B) (atbtc (c (atb)s(D) t a) (r st b) |
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Answer» Option d is correct. |
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| 21. |
Find the area of a rectangular park which is 18 m long and 83 m broad. |
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Answer» Area= breadth x length= 93/5*26/3161.2m^2 |
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| 22. |
In Fig. 7.21, AC = AE, AB = AD and<BAD= < EAC. Show that BC DE. |
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| 23. |
what is median in data handling |
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Answer» pls tell me fast and it's Formula Themedianis a simple measure of central tendency. To find themedian, we arrange the observations in order from smallest to largest value. If there is an odd number of observations, themedianis the middle value. If there is an even number of observations, themedianis the average of the two middle values. |
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| 24. |
Salve the equation |
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| 25. |
3Erample 3: Salve |
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Answer» (2x+15)/6=-3/22x+15=-18/22x=-9-15=-24x=-24/2=-12 |
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| 26. |
Salveold w |
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| 27. |
Z AB is a line segment and P is its mid-point. D andE are points on the same side of AB such thatBAD = < ABE and EPA-L DPB(see Fig. 7.22). Show that(ii) AD BE |
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Answer» EPA = ∠ DPB (Given) ⇒ ∠ EPA + ∠DPE = ∠ DPB + ∠DPE ∴ ∠APD = ∠BPE Consider Δ’s DAP and EBP AP = BP (Given P is midpoint of AB) ∠BAD = ∠ABE (Given) ∠APD = ∠BPE (Proved) Hence ΔDAP ≅ ΔEBP (By ASA congruence rule) ⇒ AD = BE (CPCT) |
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| 28. |
B is a line segment and P is its mid-point. D andE are points on the same side of AB such that4 BAD L ABE and EPA DPB(see Fig. 7.22). Show that7. A(ii) AD BE |
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| 29. |
The abscissa oh-of-low-atpointos dinate -by-3-Thenä¸14pis- |
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| 30. |
Pizzsalveusingidentitiesevalue7,2 |
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Answer» 5041 is the right answer hiii here is the answer 71^2 = ( 70 + 1 )^2 using ( a + b) ^2 = a^2+b^2 +2ab= 70^2 + 1^2 + 2×70×1=4900 + 1 + 140= 5041 answer
5041 is correct answer. 71^2=(70+1)^2(a+b)^2=a^2+b^2+2ab=70^2+1^2+2×70×1=4900+1+140=5041 5041 is the correct answer 71×71=5041 right answer 70^2=(70+1)^2using (a+b) ^2=a^2+b^2+2ab=70^2+1^2+2×70×1=4900+1+140=5041 |
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| 31. |
Ezercise 11.4In the given figure, If AB DC andPis the midpoint of BD, prove thatPis also the midpoint of AC.A. |
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Answer» Note I have used AAS congruency I forgot to write there. |
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| 32. |
\frac{1}{2} \tan ^{-1} x=\cos ^{-1}\left[\frac{1+\sqrt{1+x^{2}}}{2 \sqrt{1+x^{2}}}\right]^{2} |
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Answer» thank you |
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| 33. |
. Pis the point (-3, 4), find the sum of distances of P from both the ax |
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Answer» Distance of point P(-3,4) fromx axis = 3 Distance of point from y axis = 4 Sum of distance of P from both the axes = 3 + 4 = 7 |
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| 34. |
Salve the eguation 13-4m=23 |
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| 35. |
Salve. 3-5ct2x_-1-2 (1-x) |
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Answer» 14+2n-6+8n=4n-21n+n+348+10n=5n+135n=13-8=5n=1 3-5x+2x=-2-2(1-x)3-3x=-2-2+2x3-3x=-4+2x3+4=2x+3x7=5xx=7/5 |
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| 36. |
352 + 16 -5 X 110 = ? |
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Answer» 352+16-5×110352+16-550368-550-182 352 + 16 - 5 × 110= 352 + 16 - 550=368 - 550= -182the answer is -182 |
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| 37. |
5. Find the value of x if:(i) x% of 125 is 8(iv) x% of 110 is 11(iii) x% of 15200 is 1824(ii)1%of 1 kg is 50 g(v) x% of 320 is? 272 |
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Answer» i) x*125/100 = 8 x = 8*100/125 x = 6.4 |
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| 38. |
Salve the following egerations.i) a- 29x+100 0 |
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Answer» x^4 - 29x^2 + 100 = 0x^4 - 25x^2 - 4x^2 + 100 = 0x^2(x^2 - 25) - 4(x^2 - 25) = 0(x^2 - 4)(x^2 - 25) = 0(x - 2)(x + 2)(x - 5)(x + 5) = 0x = 2, - 2, 5, - 5 |
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| 39. |
MATHEIn Fig. 7.21, AC = AE, AB = AD andBAD =EAC. Show that BC = DE.ㄑㄧ |
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| 40. |
what is data handling |
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Answer» Data handlingis one of the central activities in which real mathematicians engage: they are frequently analysingdatathat they have gathered in various contexts and looking for patterns and generalities within them. ... So, we have adata handlingproblem that focuses on analysis rather than collection. |
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| 41. |
In the adjacent figure, AC=AE,AB=ADand <BAD=<EAC. Show thatBC = DE. |
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| 42. |
In Fig. 7.21, AC = AE, AB = AD2 BAD = EAC. Show that BC = DE.and |
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| 43. |
6. In Fig. 7.21, AC AE, AB = AD andBAD-EAC. Show that BC = DE. |
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| 44. |
120In Fig. 7.21, AC = AE, AB = ADL BAD = EAC. Show that BC = DE.and |
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| 45. |
In the adjacent figure, AC AE, AB-ADand Z BAD EAC. Show thatBC = DE. |
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| 46. |
पं) कि॥ 0 के ८०9 न 2 झान 110 + ८010 > कज 7 |
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Answer» Please like the solution 👍 ✔️👍 |
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| 47. |
\left. \begin{array} { l } { 55 ^ { \circ } , 65 ^ { \circ } , 60 ^ { \circ } } \\ { 125 ^ { \circ } , 45 ^ { \circ } , 35 ^ { \circ } } \\ { 115 ^ { \circ } , 35 ^ { \circ } , 30 ^ { \circ } } \end{array} \right. |
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Answer» option A and option C are the correct answers. The sum of angles of a triangle is 180°. therefore by adding the above angles, 55°+65°+60°= 180° and115°+35°+30 = 180° |
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| 48. |
(2*sin(68^circ))/cos(22^circ) - 2*1/(5*tan(75^circ))*cot(15^circ) - 3*tan(20^circ)*tan(40^circ)*tan(45^circ)*tan(50^circ)*tan(70^circ)/5 |
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Answer» 2 sin68/ cos 22+ 2 cot15/5tan75-3 tan45tan20tan40tan50tan70/5=2 sin(90-68)/ cos22 - 2 cot 15/5 tan(90-75)-3 tan45tan20tan40tan50tan70/5=2cos22/ cos22 - 2 cot15/5 cot15-3tan45tan20tan50tan70/5=2-2tan45tan20tan50tan70/5=2+1+1+1/5=5/5=1 1 is the correct answer of the given question |
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| 49. |
6,In Fig. 7.21, AC = AE, AB = AD andBAD = < EAC. Show that BC = DE.Flg: 72ind pis its mid-noint. D and |
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Answer» hi |
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| 50. |
acos(sqrt((1/(2*sqrt(x^2 %2B 1)))*(sqrt(x^2 %2B 1) %2B 1))) |
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