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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. |
ane in lhe20 23 then |
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Answer» The greatest angle is 90 Deg |
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| 2. |
lhe. Sunt of all terms-of an A.Ps |
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| 3. |
If tan-r-tan-x = l then lhe value of |
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| 4. |
3. ane lhe one of2solutionbyhim notionmetho |
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Answer» x+4y=8; 2x+3y=7; 3(x+4y=8); 3x+12y=24; 4(2x+3y=7); 8x+12y=28; /5x=4;:; x=4/5; ; 4/5+4y=8, 4+20y=40; 20y=36; y=36/20=9/5 |
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| 5. |
lhe Cloth in ual preces frond theof each pie ce(A |
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Answer» total length of clothe=24No. of pieces= 9therefore length of each piece= 24/9= 8/3 m = 2.67m |
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| 6. |
to be 60. Find the tme lakl16. The angles of elevation of the top of a tower from two points at a distance of 4 m and9 m from the base of the tower and in the same straight line with it are complementaryH Prove that the height of the tower is 6m.1 |
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Answer» धन्यवाद |
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| 7. |
पे J \q? Y Hrover N O( वि(Vi) a +3b)2 and4{a — 3’ t |
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| 8. |
42362oa4122 lifă495e. 33 |
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Answer» 3 a) (4^5 ×4^6 )÷4 = 4¹¹/4 = 4^10. |
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| 9. |
PAGE NO.DATEopezLHE 50 ond LB10 tind the measuieot its other tuo ongle |
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| 10. |
2. Find six rational numbers between 3 and4 |
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| 11. |
wie digit at the tens place. Find the time(iv) Vedant can row down8 km in 2 hours. Find hissspeed of the water current.In a schoolown stream 24 km in 2 hours and upstreamS. Find his speed of rowing in still water and |
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Answer» let speed of rowing =x km/hspeed of current = y km/h(x>y)down stream speed=x+y km/hupstream speed=x-y km/hspeed=dist/timex+y= 24/2=12----(1)x-y=8/2=4------(2)by adding 2x=16x=16/2=8 km/hspeed of rowing=8 km/hrputting the value of x in eq1x+y=12y=12-8=4 km/hspeed of current=4km/ht |
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| 12. |
Example 3.41Find GCD of the following:(1) 16x'y?, 24xyz |
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Answer» 4²x³y²,8*3xy³z...... 16x^3y^2, 24xy^2z= 4(4x^3y^2. 6xzy^2) 16x^3y^2,24xy^2z= 4(4x^3y^2. 6xzy^2) |
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| 13. |
7 Exercise1) If the GCD of 96 & 168 is 24. What is the LCM of these numbers |
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Answer» Given :GCD(96, 168) = 24 We know, GCD × LCM = Product of numbers24 × LCM = 96 × 168LCM = (96 × 168)/24LCM = 672 |
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| 14. |
Two numbers N and 16 has LCM of 48 and GCD of 8. what is N? |
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Answer» number1×number2=LCM×GCDN×16=48×8so N=24 |
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| 15. |
23. Solve by transposition method:-i)3x2x8+75x-253x-14iii)7x+13i)x1SX+3iv)2(x-3)-3(4-X)-1222 |
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Answer» what about 2,3and4 thanks a lot |
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| 16. |
6^(2*x %2B 4)=2^(x %2B 8)*3^(3*x) |
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| 17. |
\frac 1 3 , \frac 5 3 , \frac 9 3 , \frac 13 3 |
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| 18. |
(-8)/3 - 1/4 %2B (-11)/6 %2B 3/8 |
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Answer» last answer is -105 by24 |
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| 19. |
-1/2 %2B (5/8)/((3/4)) %2B 3/2 |
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Answer» 7/2+5/8÷ 3/4 |
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| 20. |
Ek circular object h jiska radius 9 m hai. Aur wo 2 and half times rotate karta hai.to uska displacement kya hoag |
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Answer» displacement=diameter of the circle=2×9=18cm |
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| 21. |
is 44 years,nndtmedgec9. Malini will be 6 times her present age after 30 years. How old is she now?HoTs]n certain nrize money in a cookine comnetition. She spent half the nrize |
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| 22. |
19. 65% of 7+35% of 3 - 7% of 56.ro |
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| 23. |
(X 3SOROLN(90 — ) — sec(90 — ) cos ecl+ sin® 25+ sin® 65 + /3 tan 61an 16tan 74 tan 84 |
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Answer» Cotθ.tan(90°-θ)-sec(90°-θ)cosecθ+(sin²35°+sin²55°)+√3(tan5°.tan15°.tan30°.tan75°.tan85°)=cotθ.cotθ-cosecθ.cosecθ+sin²35°+sin²(90°-35°)+√3{tan5°.tan15°.(1/√3).tan(90°-15°).tan(90°-5°)}=cot²θ-cosec²θ+sin²35°+cos²35°+√3×1/√3(tan5°tan15°cot15°cot5°)=-1+1+{tan5°tan15°(1/tan15°)(1/tan5°)}=1 |
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| 24. |
Rewrite the polynomical in standard form 7x3- x4x- |
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Answer» tq |
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| 25. |
43265/ 3 2 1+11ab cd |
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Answer» a) f(1) = 2b) lim x→1 f(x) ... is not defined as different values are there in 1+ and 1- c) f(4) is not defined... see the hole at x=4d) lim x→4 is = 2 |
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| 26. |
Q4. H.CM (2520, 6600 ] = 120 and LOM (2520,6600] =252K.Find the value of R.105. Giver that Heck [v, 65+3=1 and L.C. [1, 65+ 3 = 22338,Find u. |
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Answer» 4. 5. |
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| 27. |
2(i) x5x +6, x4x 12 whose GCD is x-2.44 |
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Answer» L.C.MxG.C.D = f (x)xg (x) f (x) = x² - 5 x + 6 g (x) = x² + 4 x - 12 G.C.D = (x - 2) L.C.M = [ f (x)xg (x) ]/G.C.D = [ (x² - 5 x + 6) (x² + 4 x - 12) ]/(x - 2) To simplify this we have to factorize these quadratic equations.So that we get = [ (x - 2) (x - 3) (x + 6) (x - 2) ]/(x - 2) = (x - 2) (x - 3) (x + 6) |
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| 28. |
-65/3 %2B 15/2 |
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| 29. |
(28)^{3}+(-15)^{3}+(-13)^{3} |
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Answer» 21952+(-3375)+(-2197)16380 will be the answer thanks fo answer |
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| 30. |
4. A. P. 1 5 9 1333161(B(C) 5(D) 191 5 9 13.-,-,-,_3 '3323is :15th term of A. P61(B) 6(A) 3(D) 19C) 5 |
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| 31. |
( 27 ) ^ { 3 } - ( 14 ) ^ { 3 } - ( 13 ) ^ { 3 } |
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Answer» thank you it is right now i am understood |
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| 32. |
(-5)^3 %2B (-8)^3 %2B 13^3 |
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Answer» 1560 the correct answer of the given question |
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| 33. |
Is (3-13) (3 +3 ) rational or irrational. |
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| 34. |
13. A flagstaff stands on the top of a 10 m hightower. From a point on the ground the angleof elevation of the top of the flagstaff is 609and that of the tower is 45 Find the distanceof the point of observation (on the ground),from the top of the flagstaff. Also find theheight of the flagstaff. |
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Answer» 7.32 m is the correct answer of the given question A|60° | | B45° | | | C------------------------------DBC= TOWERCD= DIST FROM TOWER TO PTIN∆ BCD TAN45= BC/CD1= BC/CDCD= BCIN∆ACD TAN60= AC/CD√3CD= AB+BC√3BC-BC= 10BC(√3-1)=10BC= 10(√3+1)/(√3-1)(√3-+1) = 10(1.732+1)/3-1 = 5×2.732 = 13.66mheight of tower and dist from tower to the point = 13.66m |
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| 35. |
x4x-132 =x2+12x-11x-132 |
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Answer» No solution as all the terms containing x get cancelled. |
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| 36. |
\begin{array} { l } { \text { Divide : } } \\ { \text { (i) } 132 \text { by } 11 } \\ { \text { (iii) } 132 \text { by } ( - 11 ) } \end{array} |
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Answer» i)132÷11=132/11=12ii)132÷(-11)=132/(-11)=-12 |
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| 37. |
122 '6 |
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Answer» thank you |
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| 38. |
122% of 800 |
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Answer» (12*2+1/2)/100 *800 (25/2)*8 =25*4=100 |
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| 39. |
122%of 400 |
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| 40. |
Find x #, *21-122 |
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Answer» x² - 4 = -4x - 4x² + 4x + 4 - 4=0x² + 4x = 0 x² = - 4x x² / x = -4 x = -4 x= -4 is the best answer x=-4 is the correct answer x2-4=-4x-4x2+4x+4-4=0x2=-4xx2/x=-4x=-4 x= -4 is the correct answer of the given question |
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| 41. |
Simplify (3 +32(2+122) |
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| 42. |
3) A triangle has a base of 10 m. If the height is 2x m and the areathe area and the height of the triangleit the height is 2x m and the area is 80x sq. m. find |
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Answer» Area of triangle = 1/2 * base * height 80x^2 = 1/2*10*2x 80x = 10 x = 10/80 = 1/8 Therefore,Area = 80*(1/8)^2 = 10/8 = 5/4 = 1.25 m^2 Height = 2x = 2/8 = 1/4 =.25 m |
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| 43. |
Ataparticulartimeofaday, a 7-m-high flagstaff casts a shadow which is 8.2 m long. What isthe height of the building which casts a shadow 20.5 metres in length at the same moment? |
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| 44. |
At a particular time of a day, a 7-m-high flagstaff casts a shadow which is 8.2 m long What isthe height of the building which casts a shadow 20.5 metres in length at the same moment? |
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| 45. |
A committee of 6 is to be chosen from 10 persons with the condition that if a particular pechosen, then another particular person B must be chosen. |
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Answer» here 2 persons are fixed so we have to select other 4 from remaining 8 personsso 4/8=1/2 wrong answer |
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| 46. |
In how many ways can a cricket teamof 11 be chosen out of a group of 15players? How many of them will(a) include a particular player?(b) exclude a particular player?4 |
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| 47. |
x+4=67-y y-6=80xfind the value of x |
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Answer» And y = 63-(57/81)62.29 |
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| 48. |
i buys a fish from a shop for his aquarium. Theshopkeeper takes out one fish at random from atank containing 5 male fish and 8 female fish (seeFig. 15.4). What is the probability that the fish taken1l. Gopeout is a male fish? |
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| 49. |
6)Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from atank containing 5 male fish and 8 female fish.What is the probability that the fish taken out is a malefish? |
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| 50. |
; b2lle 124 i) lbalbha 01 b लक 5 220६ e k 0T URpihS lmm lalelk ralphe 01rEih Jalp|e lkalkhd 5 ‘2ue 05६ 2 Rhul 10D Yal2lb2lte 12|k lkalkhd 125 p lhalkha § |3l 221 292010 14h - |
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