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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. |
\left(\frac{5^{-3} \times 3^{4}}{3^{-2}}\right) \div \frac{27}{125} |
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| 2. |
225*a^2 - 135*a - 125*a^3 %2B 27 |
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| 3. |
the trial and error3) 5y-3-12 |
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Answer» in a table |
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| 4. |
3) 14-K = 2k + 4y the trial and error4 3) 5y-3-12 |
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Answer» 5y-3= 125y= 15y= 3thanksplease like the solution 👍 ✔️ |
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| 5. |
66. The sales price of an item was decided so that there is 25% profit. If 12then what should be the discount the decided price? [ Ans : 10%)% profit is to be earned |
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Answer» Let the price be rs 100 then to get 25% profit it should be marked as 125 rs.To have 12.5 % it should be sold at 112.5Discount % = (125-112.5)*100/125 = 10% |
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| 6. |
Radhika decided to donate 18% of her monthly salary to an NGO. On theday of donation she changed her mind and donated7749 which was 82%of what she had decided earlier. How much is Radhika's monthlysalary?(B) 52500(A) 42800(E) of these(C)757500(D)56500 |
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| 7. |
4. A ratio in its lowest terms in 3: 7. If the difference between them is 16, find the quantitiesiwai 2014) IAns. 12 and 28 |
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Answer» let the quantities be 3x and 7x 7x-3x =164x=16 x=4quantities are 12 and 28 |
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| 8. |
CBSE2011)A A straight highway leads to the foot of a tower. A man standing on thetop of the tower observes a car at an angle of depression of 30°, which isapproaching the foot of the tower with a uniform speed. Six secondslater, the angle of depression of the car is found to be 60. Find the timetaken by the car to reach the foot of the tower form this point.HINT See solved Example 14.] |
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| 9. |
w taps running together can fill a bath in 4 minutes, which is filled by one of the taps by itsen7 minutes. How long would it take if the other pipe is running by itself? |
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Answer» tank filled by both tap in 4 minute= 1 tank filled by both tap in one minute= 1/4 tank filled by one tap in 7 minute= 1 tank filled by one tap in 1 minute = 1/7 let tank filled by other tank in 1 minute be x 1/7 +x = 1/4 x = 1/4 - 1/7 x = 7/28 - 4/28 x = 3/28 tank filled by other tap in one minute = 3/28 3/28 tank is filled by tap in =1 min 1 tank is filled by tap in = 28/3. min 28/3 minutes is the right answer |
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| 10. |
What length of 5m wide cloth will be required to make a cowhose base is 7m and whose height is 24m? (Ď-)22//- |
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Answer» thanks |
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| 11. |
440m. Find the width and area 01 thA 3 m wide road isall around on the outside of a 125 m long and 60 m wide garden. Findasrepairing the road at the rate of Rs. 150 m2theSummary(a) F |
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Answer» Area of rectangular park = length × breadth= (125 × 60) m^2= 7500 m^2Since 3m wide path runs outside the park .So the length of park along with path is 131m and the breadth of park along with park is 66.Area of park with path = 131×66= 8646m^2Area of path = area of rectangular park with path - area of park= (8646-7500)m^21146m^2= 1146 m^2cost will be1146*150171900 ruppes wrong answer what is the answer for thissorry for the mistakes |
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| 12. |
respectively. Find the distance between the ears. iakeICBSE 2011)14) A straight highway leads to the foot of a tower. A man standing on thetop of the tower observes a car at an angle of depression of 30°, which isapproaching the foot of the tower with a uniform speed. Six secondslater, the angle of depression of the car is found to be 600. Find the timetaken by the car to reach the foot of the tower form this point.HINT See solved Example 14.] |
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| 13. |
Example 6 : In Δ ABC, right-angled at B,AB = 5 cm and ACB = 30° (see Fig. 8.19).Determine the lengths of the sides BC and AC. |
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| 14. |
[If the co-ordinates ofa variable point P be (cos θ + sin θ, sin θ-cos0), where θ is a variable quantity, find the locus of P.] |
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Answer» P = (x1, y1) = (cosx+sinx, sinx-cosx) x1 = cosx+sinxy1 = sinx-cosx , quadr the eq's => (x1)² = (cosx+sinx)² = cos²x + sin²x + 2sinxcosx(y1)² = (sinx-cosx)² = sin²x + cos²x - 2sinxcosx , add the eq's (x1)² + (y1)² = 2It's a circle with center (0, 0) and radius √2 |
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| 15. |
Example 6: A corn cob (see Fig. 13.17), shaped somewhatlike a cone, has the radius of its broadest end as 2.1 cm alength (height) as 20 cm. If each 1 cm2 of the surface of thecob carries an average of four grains, find how many grainsyou would find on the entire cob.nd |
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Answer» Since the grains of corn are found on the curved surface of the corn cob. Total number of grains on the corn cob = Curved surface area of the corn cob x Number of grains of corn on 1cm^2 Now,find the curved surface area of the corn cob.r = 2.1 cmh = 20 cmLet l be the slant height of the corn cob. Then,l^2 = r^2+ h^2= (2.1)^2 + (20)^2=4.41 + 400=404.41l =20.11 C.S.A. of cone= πrl= 22/7 * 2.1 * 20.11= 132.73 cm^2 We know that, Total no. of grains on the corn cob= 132.73 * 4= 530.92 so, approximately there are 531 grains. |
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| 16. |
Q.1The general solution ofthe equation tan2α+2V3 tan α-l is given by(2n + 1) Tl(6n + 1) π(D) α_nT2(A) α= 2(B) α(C) α |
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| 17. |
each question is followed by the values of variable decided whether these values are the solution of thatequation X-4=3, x= 1,7,-7 |
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| 18. |
10. A wire, when bent in the form of a square,encloses an area of 484 cm2. Find:(i)one side of the square |
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| 19. |
A steel wire when bent in the form of a square encloses an area of 121 cm2. The same wireis bent in the form of a circle. Find the area of the circle. |
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| 20. |
find the sides of the two squares.Find the area and the perimeter of a squarewith diagonal 24 cm.[Take-/2 = 1-41].A steel wire, when bent in the form of asquare, encloses an area of 121 cm, The6 |
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Answer» thank u bro |
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| 21. |
& A copper wire when bent in the form of a square encloses an area of 121 cm2. If thesame wire is bent into the form of a circle, find the area of the circle. |
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| 22. |
15. Two paths each of3 m width are running perpendicular120n+pABto each other in the centre of a square park of side 120 m.0-Find the total area of the pathD C |
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Answer» Area of path=3*120+3*120-3*3=360+360-9=720-9=711m^2 |
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| 23. |
0-5-5)m391,5 m2, of construction of roads391.5x750293,625.0Exercise 11.4arden is 90 m long and 75 m broad. A path 5m wide is to be built outside and around it. Findof the path. Also find the area of the garden in hectare. |
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| 24. |
7. A rectangular ground 80 m x 50 m has a path 1 m wide outside aroundit. The area of the path is |
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| 25. |
꾏) A trce is bwken by the uindTF the Point.fyom where it broke 15 ia m above the qoundand its top touches the ground at a distan35e Doihthee be Poye it brokeToKe |
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| 26. |
2.How will an organism be benefited if it reproduces through spores? |
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| 27. |
(vi) ΔABCCD is a trapezium in which AB II CD andAD- LA BBC (see Fig. 8.23). Show that: E(iii) Δ ABC Δ BAD(iv) diagonal AC diagonal BDFig. 8.23Hint: Extend AB. and draw a line through Cparallel to DA intersecting AB produced at E.) |
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| 28. |
CD is a trapezium in which AB || CD and A-BC (see Fig. 8.23). Show thatB.FEAABCE A BADdiagonal AC diagonal BDint: Extend AB and draw a line through CFig. 8.23parallel to DA intersecting AB produced at E.] |
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| 29. |
Add and 79 |
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Answer» Add (-5/9) and (-17/9) Since denominator is same for both fractions = (- 5 - 17)/9= - 22/9 ans |
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| 30. |
7.Fig.ABCD is a rhombus. Show that diagonal ACbisects Z A as well as Z C and diagonal BDbisects < B as well as Z D. |
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| 31. |
ABCD is a trapezium in which AB II CD andAD BC (see Fig. 8.23). Show that(iii) AABC ABAD(iv) diagonal AC- diagonal BD |
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| 32. |
ABCD is a trapezium in which AB iICIDand AD BC (see Fig. 7.23). Show that(ii) < C = < D(iii) Î ABC Î BAD(iv) diagonal AC diagonal BD |
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Answer» i) AB||CD(given) AD||EC (by construction) So ,ADCE is a parallelogram CE = AD(Opposite sides of a parallelogram) AD = BC (Given) We know that , ∠A+∠E= 180° [interiorangles on the same side of the transversal AE] ∠E= 180° - ∠A Also, BC = CE∠E = ∠CBE= 180° -∠A ∠ABC= 180° - ∠CBE [ABE is a straight line] ∠ABC= 180° - (180°-∠A) ∠ABC= 180° - 180°+∠A ∠B= ∠A………(i) (ii) ∠A+∠D = ∠B+∠C = 180° (Angles onthe same side of transversal) ∠A + ∠D = ∠A + ∠C (∠A = ∠B) from eq (i) ∠D = ∠C (iii) In ΔABC and ΔBAD, AB = AB (Common)∠DBA = ∠CBA(from eq (i) AD = BC (Given) ΔABC ≅ ΔBAD (by SAS congruence rule) (iv)Diagonal AC = diagonal BD (by CPCT as ΔABC ≅ ΔBAD) |
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| 33. |
(vi) Δ ABC Δ DEF.ABCD is a trapezium in which AB II CAD+ BC (see Fig. 8.23). Show that(ii) KABCEA BAD(iv) diagonal AC diagonal BD |
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Answer» Construction: Draw a line through C parallel to DA intersecting AB produced at E. Proof: i) AB||CD(given) AD||EC (by construction) So ,ADCE is a parallelogram CE = AD (Opposite sides of a parallelogram) AD = BC (Given) We know that , ∠A+∠E= 180° [interior angles on the same side of the transversal AE] ∠E= 180° - ∠A Also, BC = CE ∠E = ∠CBE= 180° -∠A ∠ABC= 180° - ∠CBE [ABE is a straight line] ∠ABC= 180° - (180°-∠A) ∠ABC= 180° - 180°+∠A ∠B= ∠A………(i) (ii) ∠A+∠D = ∠B+∠C = 180° (Angles on the same side of transversal) ∠A + ∠D = ∠A + ∠C (∠A = ∠B) from eq (i) ∠D = ∠C (iii) In ΔABC and ΔBAD, AB = AB (Common) ∠DBA = ∠CBA(from eq (i) AD = BC (Given) ΔABC ≅ ΔBAD (by SAS congruence rule) (iv)Diagonal AC = diagonal BD (by CPCT as ΔABC ≅ ΔBAD) |
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| 34. |
1. A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outsidearound it. Find the area of the path. Also find the area of the garden in hectare T· |
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| 35. |
garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside andaround it. Find the area of the path. Also find the area of the garden in hectare |
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| 36. |
.A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outsideits border. Find the area of the path. |
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Answer» thanks |
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| 37. |
A garden is 90 m long and 75 m broad. A path 5 m wide is to bearound it. Find the area of the path. Also find the area of the garden in hectarebuilt outside1.nd |
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Answer» woo |
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| 38. |
1. A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outsidearound it. Find the area of the path. Also find the area of the garden in hectare.Dand4cm |
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| 39. |
A garden is 90 m long and 75 m broad. A path 5 m wide is to be builaround it. Find the area of the path. Also find the area of the garden in hectare f1.t outside |
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| 40. |
4. Divide 20 pens between Sheela and Sangeeta in the ratio of 3:2 |
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| 41. |
L4 Divide 20 pens between Sheela and Sangeeta in the ratio of 3:2. |
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Answer» ratio 3:2let x be common ratiothen 3x+2x=205x=20x=4no.s are 12 and 8 |
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| 42. |
What change should be made on LHSof an equation when we add 7 to theRHS? |
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Answer» 7 should be added to LHS also so that both LHS and RHS are balanced 7 should be added to LHS 7 add on LHS to make the equation equal |
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| 43. |
2.7 Equations Redu2x +3Solution: Observe that the equLHS is t at But we can pu |
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| 44. |
7. The base of the rectangle is 22 cm and the area is 484 cm2, what is the height of the rectangle? |
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| 45. |
15. The cost of fencing a square gardenis 400 at the rate of t per metre. Find the area of the garden in squaremetres. |
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| 46. |
A book consists of 264 pages payal had read 3/4 of the book till yesterday how many more pages she has to read? |
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Answer» read pages = 264 * 3/4 = 198 remaining pages = 264-198 = 66 pages Remaining pages in fraction =1/4so in number it is 1/4*264, which equals to 66 pages more to read |
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| 47. |
Neha covered 95.52 m by running around a square garden. Find the length of each sideof the garden. |
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Answer» length of one side of a square = = 95.52/4 = 23.88 |
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| 48. |
a covered 95.52 m by running around a square garden. Find the length of each sideof the garden. |
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Answer» Perimeter square is 95.52side4×s=95.52s= 95.52/4s=23.88m |
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| 49. |
3. The cost of fencing a square garden at therate of 5 per metre is 2,000. Find thearea of the garden. |
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Answer» area will be 400m²per metre charges = 5 rupeesor we can say in 5 rupees - 1 metreso in 1 rupee= 1/5 metreso in 2000rupees = 1/5*2000 = 400m² |
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| 50. |
a rectangular garden is surrounded on its outside by 2 meter broad path. if the area of the path is 464 square metres. then what is the parameter of the garden |
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