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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. |
TThe shape of agarden isrectangular in20 m -the middle and semi circular at the endsas shown in the diagram. Find the areaand the perimeter of this garden (Lengthof rectangle is 20-(3.5 +3.5) metres). |
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| 2. |
Fig. 11.0in Fig. 11.41, if AB II DE and BD II FG such that ZFGH = 125° and <B = 55°, fand y8.find, M2.55。125Fig. 11.41 |
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Answer» thanks |
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| 3. |
100-120 |
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Answer» Ans :- 100 - 120 = -20 |
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| 4. |
4 ^ { \circ } + \operatorname { cos } 55 ^ { \circ } + \operatorname { cos } 125 ^ { \circ } + \operatorname { cos } 204 ^ { \circ } + \operatorname { cos } 300 ^ { \circ } = \frac { 1 } { 2 } |
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| 5. |
\frac 14 \times 4.5 1.4 \times 0.45 = |
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Answer» 14×4.5 ÷1.4× 0.45 63÷0.63 100 |
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| 6. |
\frac { 2 } { 5 } \times ( - \frac { 3 } { 7 } ) - \frac { 1 } { 6 } \times \frac { 3 } { 2 } + \frac { 1 } { 14 } \times \frac { 2 } { 5 } |
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| 7. |
Exercis8is given that dABCARPQ. Is it true to say that BC = QR? Wh |
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Answer» If you find this solution helpful, Please give it a 👍 |
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| 8. |
\frac{2}{5} \times\left(-\frac{3}{7}\right)-\frac{1}{6} \times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5} |
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| 9. |
Simplify the followin1.89415 +35 |
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Answer» 10/9+7/2=83/18=3/5*83/1883/30 |
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| 10. |
5. It is given that AABC APQR with BC: QR 1:3. Find the ratio of area of APRQ to areaof ABCA |
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Answer» ABC ~ PQR Area of ABC / Area of PQR = BC*BC/QR*QR = 1/9 |
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| 11. |
26, A LABC is given. If lines are drawn through A, B,QAC, parallel respectively to the sides BC, CA and AB,forming Δ POR, as shown in the adjoining figure, showthat BC =-QR. |
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| 12. |
6. In Fig. 6.21, A, B and C are points on OP, OQ and|| PR.OR respectively such that AB I| PQ and ACShow that BC| QR. |
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| 13. |
15 \div 3 \times 7 + 14 - 3 \times ( 1 + 5 ) |
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| 14. |
(4)0.45The value of 025.0.22. is(B)014(C)0.45(D)OAThe value of 2x2 is(16)2(C)3(D) of these16/13-9/52 is equal to(D) of theseis an(A) natural number (B) rational number(D) Irrational number(C) integer2+5 55 2236 and V10 - 3.162Find the value ofConvert 0.25 into rationalSimplify (3,15 + 2/2)(2-85 +32)SimplifySimplify W28-11-2310. 12-0142857. And the value of 1411. Find 6 rational no. between and12.Show how va can be represented on the number lineFind a and bif 3-16=avc-b3+2/6 |
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Answer» ans-1161/5,62/5,63/5,64/5,66/5,69/5 |
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| 15. |
13. The length of a rectangular plotis twice of its width. If the lengthof a diagonal is 9V5 metres, theperimeter of the rectangle is |
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| 16. |
Inform 0.35 is(A) 35/9(B)35/99 (C) 315/99 (D) 35/10 |
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Answer» x=0.353535...this is equation 1 multiply with 100 100x = 35.353535...this is equation 2 equation 2 - equation 1 100x = 35.3535... x = 0.3535...________________99x = 35 x = 35/99 35/99 is the answer.... |
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| 17. |
(a) 35 (b) 40 (c) 45 (d) 55(a) 80 (b) 100 () 120 (d) 90 |
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Answer» Sorry we are currently taking questions in English only. please solve this question |
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| 18. |
A.105B.35C.210D.15Ă21Ă35 |
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Answer» Explanation : 15 = 3×5 21 = 3×7 35 = 5×7 લ.સા.અ. = 3×5×7 = 105 Solution : 105 If you find this answer helpful then like it. |
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| 19. |
A quarter of a circle is called........... |
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Answer» Ans :- A quarter of a circleor circular body, viz. (a) an arc of acircle, forming one fourth of the circumference; (b) one fourth of the area of acircle, contained within two radii at right angles. A thing having the form ofa quarter-circle. |
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| 20. |
rectangular sheet of paper is 15-cm long and 12cm wide. Find i |
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Answer» right |
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| 21. |
\frac { 5 ^ { 2 ( n + 6 ) } \times ( 25 ) ^ { 2 n - 7 } } { ( 125 ) ^ { 2 n } } |
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Answer» = 5^2(n+6) * (5^2)^2n-7/ (5^3)^2n = 5^ 2n+12+4n-14 / 5^6nAs bases are same, powers can be equated = 6 n - 2 - 6 n = -2. |
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| 22. |
the altitude AD in a triangle ABC is 12cm long. If BD is 8cm and DC is 18cm, prove that the angleBAC is a right angle |
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| 23. |
In Fig. 6.21, A, B and C are points on OP, OQ andOR respectively such that AB | PQ and AC|| PR.Show that BC| QR. |
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| 24. |
In Fig. 6.21, A, B and C are points on OP, OQ andOR respectively such that AB PQ and AC || PR.Show that BC | QR. |
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| 25. |
The L.С.М of two numbers is 14 times their H.C_F, the sum of L.С.М andН.С.F is 600. If one number is 280 then find the other number. |
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Answer» Let the H.C.F be xLet the L.C.M = 14xx + 14x = 60015x = 600 X = 40HCF = 40LCM = 14 X 40 = 560Other number =560×40/280=80 |
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| 26. |
6. In Fig. 6.21, A, B and C are points on OP, OQ and EOR respectively such that AB || PQ and AC || PR.Show that BC||QR. |
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| 27. |
.280 काकिरा होता है। |
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Answer» 280x65/100=28×650=18,200 65% of 280=65*280/100=18200 is the right answer 18200 is the answer of the following 280×65/100182is correct answer the correct answer is 18200 of the following question 18200 is the right answer 280 x 65/100=18200/100=182 280×65/100=182 is the answer of the given question. 18200 is correct answer 18200 is the right answer 18200 is right answer correct answer is 182 182 is the right answer 280 ka 65%182answer 182 is the correct answer 280×65%= 280×65/100=182. |
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| 28. |
. The 1.c.m. of two numbers is 14 times theirg.c.d. is 600g.c.d. and the sum of 1.c.m. andIf one of the numbers is 280, then find theother number. |
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Answer» PLEASE LIKE IT, IF YOU FIND THIS SOLUTION HELPFUL. |
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| 29. |
1.Find x in the following figures.125°ro125° |
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| 30. |
given that LCM(45, 125) =1125, find HCF(45, 125) |
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Answer» We know HCF × LCM = product of numbers 1125 × HCF = 45 × 125 HCF = (45×125)/1125 HCF = 5 |
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| 31. |
49. In the given figureAOC is a diameterand AC is parallel toED, If angle CBE = 64°,calculate DEC |
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| 32. |
2 + 7 + 12 + \dots + ( 5 n - 3 ) = \frac { n ( 5 n - 1 ) } { 2 } |
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| 33. |
SECTION-BIn given figure O is centre of the circle if <AOC = 130° then find LABC |
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Answer» ABC = half of AOC (angle substended by an arc is half the angle substended by an arc at the centre )=130÷2=65 |
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| 34. |
(A) Attempt any TWO of the following:0) Find accumulated value after 1 year of annuity immediate in which 20000 isinvested every quarter, at 16% pa, compounded quarterly.Given: (1.04) 1.1699) |
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Answer» Number of quarters = 12/3 = 4. Amount = P*[1+(r/100)]^n= 20000*[1+(16/100)]^4= 20000*[29/25]^4= ₹ 36212.78 Please hit the like button if this helped you. |
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| 35. |
Simplify and find the producta. 35 b. 74 |
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Answer» 1.3^5= 2432.7^4=2401is the best answer |
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| 36. |
Find the mode from the following data:125, 175, 225, 125, 225, 175, 325, 125, 375225, 125Q.16 |
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| 37. |
5. I P,O. R are three points on a line and Q is between P and R, then prove thatPR -QR PO. |
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| 38. |
ofD E and F are respectiveD.are respectively the mid-points of the sides BC, CA and AB of a Δ ABC.show thatBDEF is a parallelogram.(11) ar (DEF) =ar (ABC)4ii) ar(BDEF)=-ar (ABC)( |
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| 39. |
PO is a line segment 12cm long and R is a point in is interior such that PR 8cm.PQ2-PR2 and PR2+ QR2 2PR. QR. |
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| 40. |
1. કિંમત શોધો :| (i) 26 eg (ii) 93(iii) 112(iv) 54 |
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Answer» 1) 64 2) 729 3)1331 4)625 |
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| 41. |
6. PQRS is a parallelogram in which ZPSR- 125°. FindZRQT125°Q T |
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| 42. |
125+125 |
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Answer» 125 + 125 = 250 |
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| 43. |
\frac { ( 75 \times 280 ) + ( 125 \times 150 ) } { 75 + 125 } |
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Answer» 75*280=21000125+150=18750125+75=20021000+18750/200=39750/200ans=198.75 Nice answer |
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| 44. |
8. In the given figure, AOB is a straight line and the rays OC and ODstand on it.If ZAOC = 65°, ZBOD = 70° and ZCOD = xº, find the value of x. |
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Answer» answer of this question |
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| 45. |
If 52x+125 125, find the value of x. |
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| 46. |
reIn the given figure, AB = BC = CD andZABC 132°. Calculate:(i) ZAEB,(ii) <AED,(ii) ZCODveof132° |
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| 47. |
Lines and Angles3. In the adjoining figure, AOB is a straight line.Find the value of x. Hence, find ZAOC, ZCODand ZBOD(3x + |
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| 48. |
4.) In the given figure AOC is a line, find the value of2x |
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Answer» We know angle AOB + angle BOC = 180° Reason : Because it is straight line. angle AOB + angle BOC = 180° 70° + 2x = 180° 2x = 180° - 70° 2x = 110° x = 110° x = 110°/2 x = 55° Please hit like if you find solution useful |
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| 49. |
SECTION - Bउपयुक्त सातभा का प्रयोग करके 100ते ।106 usiog appropriate identityकीजिए।Evaluate 105P, 10 |
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Answer» ok 105×106= (100+5)× (100+6)(x+a)(x+b)=x^2+ (a+b)x +ab =100^2 +(5+6)(100) +(5×6) =10000 + 100(11) + 30 =10000 + 1100 + 30 =11130 |
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| 50. |
04Reted byowMOREPEN LABORATORIES LTD20 000For WUND(CETICO1000odComposition:HOMENT-171244 BFG.12/2082 ERRS.40.00 PER LO TABS.IRAO |
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