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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. |
D and E are points on AB andAC respectively of AABC, suchthat ar (ADBC) ar (AEBC). Prove that DE II BC |
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| 2. |
6. A line segment DE is drawn parallel tobase BC of A ABC which cuts AB at point Dand AC at point E. If AB 5 BD andEC-32 cm, find the length of AE. |
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| 3. |
22. Latika wants to put a border around herbedsheet of length 10 m and breadth5 m 60 cm. Find the total cost of the borderrequired at the rate of 90 per metre.(A) 2808(C) 2408(B) 2505(D) 2605 |
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Answer» Perimeter = 2×(l+b) =2×(10m+5m 60cm) =2×(15m 60cm) =2×(1500cm+60cm) =2×(1560cm) =3120cm change into meter: 31.2m cost of border per meter = Rs.90 cost of border 31.2m = Rs.31.2 × 90 = Rs2808 |
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| 4. |
Dabud & co. Ordered for 22 bicycleIf one bicycle coste P_2165. Hecalculate the amount that has tobe paid by the company. |
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Answer» Cost of 1 bicycle =2165Cost of 38 bicycle =38*2165=82270 rs 1 bycycle=2165.70= 38 x 2165.70=82296.6 Rs 38×2165.70=8229.60 cost of one cycle = 8229.90 1 bicycle= 2165.70=38×2165.70=82296.6 RS |
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| 5. |
5. Tarun can paint th of a painting in 8 days. How many days will he take to complete the paintin |
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| 6. |
"Example 9.9 : AM and AN are the altitude of ABCD. Theircorresponding base are BC are CD respectively. If AM10 units,AN = 8 units and BC = 16 units. Find CD and the Area of i AMCDnm |
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Answer» Area(A) = Base (b) x height (h)A = BC (10) x AM (16)= 160cmsq.So..CD as base,CD (b) = A/h = 160/8(AN)= 20 cm . |
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| 7. |
0-25.की, 0/2/“ |
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Answer» 2x²-7x-15=2x²-10x+3x-15=2x(x-5)+3(x-5)=(2x+3)(x-5) |
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| 8. |
I.A shirt with marked price550 is sold at a discount of 15%. Find its selling price. |
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| 9. |
Rahul bought a shirt for ₹405.85 and jeans for ₹425.50. How much did he has to pay? |
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Answer» total Rahul has to pay = price of a shirt + price of a jeans = 405.85 + 425.50 = 831.35 |
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| 10. |
lexercise 3.2iswer the following:Rahul bought a mobile for500 and soind the percentage of profit earned by Rahud sold it |
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Answer» CP of mobile = 500 SP of mobile = 625 Profit = SP - CP = 625 - 500 = 125 Profit % = profit/cp * 100 125/500 * 100 = 25% The profit% is 25% |
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| 11. |
5. ABC and DBC are two isosceles triangles on thesame base BC (see Fig. 7.33). Show that |
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| 12. |
Latika wants to put a border around herbedsheet of length 10 m and breadth5 m 60 cm. Find the total cost of the borderrequired at the rate of 90 per metre.(A) 2808(C) 2408(B) 2505(D) 2605 |
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Answer» perimeter = 2×(l+b) =2×(10m+5m 60cm) =2×(15m 60cm) =2×(1500cm+60cm) =2×(1560cm) =3120cm change into meter: 31.2m cost of border per meter = Rs.90 cost of border 31.2m = Rs.31.2 × 90 = Rs2808 thanks |
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| 13. |
are two isosceles triangles on thesame base BC (see Fig. 733). Show thass. ABC and DBC |
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| 14. |
ABC and DBC are two isosceles triangles on thesame base BC (see Fig. 7.33). Show that |
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| 15. |
ABC and DBCsame base BC (see Fig.are two isosceles triangles on the7.33). Show thatFig |
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| 16. |
5. ABC and DBC are two isosceles triangles an thesame base BC (see Fig. 7.33). Show thr |
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| 17. |
5 ABC and DBC are two isoscelestrianglesonthe. same base BC (see Fig. 7.33). Show that |
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Answer» As angle ABC = angle ACB.....(1 angle DBC = angle DCA .....(2adding (1 and (2we get ABC+ DBC = ACB + DCAso ABD = ACD , proved |
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| 18. |
Find the distance of the point (3, -5) from the line 3x-4y-26=0 |
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| 19. |
A cupboard was bought for 7080 after getting a discount of t 420. What is thepercentage of discount? |
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Answer» Marked price of cupboard = 7080 + 420= 7500Let p is percentage of discount420 = p x 7500/100p = 420/75p = 5.6Discount percentage is 5.6% |
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| 20. |
Point (-3, 5) lies in the |
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Answer» In point (-3, 5), x-coordinate is negative and y-coordinate is positive. So, the point lies in the second quadrant. |
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| 21. |
EXAMPLE 2.16: Whicat 2 m sa greater force, accelerating a 2 kg mass at 5 m s2 or a 4 kg mass |
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Answer» thnx |
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| 22. |
A cupboard was bought for rupee 7080 after getting a discount of rupee 420 . what is the percentage of discount ? |
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Answer» Marked Price = 7080+ 420 = 7500.Discount Percent = (420/7500)*100= 420/75= 5.6 % |
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| 23. |
cos 38° cos 52° —sin 38° sin |
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Answer» We know,cosA cosB - sinA sinB = cos(A+B) Here we have A = 38° and B = 52° cos(38°)cos(52°) - sin(38°)sin(52°)= cos(38°+52°)= cos(90°)= 0 |
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| 24. |
) Point (-3, 5) lies in the |
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Answer» -3 is in second quadrant and 5 is in first quadrant, So, (-3,5) is in second quadrant. |
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| 25. |
Write the distance of the point (3, -5, 12) from x-axis |
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| 26. |
3 Tarun bought a TV with 20% discount on thelabeled price. Had he bought it with 25%discount, he would have saved 500" At whatAnsw |
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Answer» Let lebeled price was 100. Tarun bought = 100 - 20% of 100 = 80. If Tarun bought it at 25% discount, = 100 - 25% of 100 = 75. Difference = 500 80 - 75 = 500. 5 = 500 1 = 500/5 80 = 100 *80 = 8000.Correct Answer is Rs. 8000. Alternatively, Difference = 500 5% = 500. 100% = (500 *100)/5 = 10000. 100% is the lebeled price. So, TV price = 10000 - 20% of 10000 = 8000. Note: You can solve it by taking lebeled price as X as well. |
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| 27. |
bought a T.V. with 20% discount on the1 Tarulabeled price. Had he bought it with 25%discount, he would have saved 500. At whatAnswe |
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Answer» Let lebeled price was 100. Tarun bought = 100 - 20% of 100 = 80. If Tarun bought it at 25% discount, = 100 - 25% of 100 = 75. Difference = 500 80 - 75 = 500. 5 = 500 1 = 500/5 80 = 100 *80 = 8000.Correct Answer is Rs. 8000. Alternatively, Difference = 500 5% = 500. 100% = (500 *100)/5 = 10000. 100% is the lebeled price. So, TV price = 10000 - 20% of 10000 = 8000. Note: You can solve it by taking lebeled price as X as well. |
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| 28. |
Kunal bought a shirt marked for 1,500 in a sale. There was 20% discount on each item and another25% discount over and above the existing discount. How much did Kunal pay for the shirt6. |
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| 29. |
38. A power trailer was bought for1,44,000 getting 10% discount. What wouldbe its price, if the discount is 8%?0% |
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Answer» Suppose marked price is xthen x-0. 1x= 1440000.9x= 144000x= 160000now 8 percentage of discount = 160000*8/10012800160000-12800= 147200rupees |
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| 30. |
13 Rahul bought a sweater and saved 20 when a discount of 25%wasgiven. What was the price of the sweater before the discount? |
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Answer» Let,Before Discount amount beRs 100,and After Discount amount is Rs 75. So, solving by Unitary Method,Rs 25 is saved when the amount is Rs 100,Re 1 is saved when the amount is Rs100/25Rs 25 is saved when the amount is Rs100/25× 20 = Rs 80 |
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| 31. |
Four friends make some punch for a.party. The amount of punch that eachmade is shown in the table below. Whichfriend made the least amount of punch?: Friend Amouit of PunchFrederico 6.25 pintsKeith2 quartsRuth0.75 gallonOliver13.5 cupsFredericoB. KeithC. RuthD..Oliyer |
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Answer» According to this9/4 quarts will be approx2.54 litrewhixh is the lowest Ruth made tge least amount |
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| 32. |
(iv) The taxi charges in a city consist of a fixed charge together with105distance covered. For a distance of 10 km the charge paid isjourney of 15 km, the charge paid is 155. What are the fixed charges.charge per km? How much does a person have to pay for travelling25 km?e together with the charge for the105 and for aharge paid isthe fixed charges and theay for travelling a distance of |
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Answer» 10 km=105rsCharge per km=105/10=10.5rs25 km charge=10.5*25=rs 262.5 Let the fixed charge be Rs x and per km charge be Rs y.According to the given information,x + 10y = 105(1)x + 15y = 155(2)From (3), we obtainx = 105 – 10y(3)Substituting this in equation (2), we obtain105 – 10y + 15y = 1555y = 50y = 10 (4)Putting this in equation (3), we obtainx = 105 – 10 × 10x = 5Hence, fixed charge = Rs 5And per km charge = Rs 10Charge for 25 km = x + 25y= 5 + 250 = Rs 255 |
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| 33. |
MATHEMATICThe taxi charges in a city consist of a fixed charge together with the charge for thedistance covered. For a distance of 10 km, the charge paid is ? 105 and forjourney of 15 km, the charge paid is 155. What are the fixed charges and thecharge per km? How much does a person have to pay for travelling a distance of25km? |
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| 34. |
22. In fig. 4, AP L BC. If AP2BP. PC, prove thatBAC 90°Fig. 4 |
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Answer» Like if you find it useful |
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| 35. |
The taxi charges in a city consist of a fixed charge together with the charge for thedistance covered. For a distance of 10 km, the charge paid is ? 105 and for ajourney of 15 km, the charge paid is 155. What are the fixed charges and thecharge per km? How much does a25 km?person have to pay for travelling a distance of |
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Answer» thank you bahi app apna whatshapp number De do |
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| 36. |
PAIRaxi charges in a city consist ofa fixed charge together with the charge for ththe charge paid is Rs 155. What are the fixed charges and thecharge per km? How much does a person have to pay for travelling a distance odistance covered. For a distance of 10 kmi, the charge paid is Rs 105 and fora So.journey of 15 km, of tVer0 |
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| 37. |
(i) In the given figure PA 6, PB= 4 and PC = 8, FAすープB |
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| 38. |
out usingFig. Q. 13Fig: Q. 12BOC+ ZBOD338°, find all the four angles.13. In the given figure, if ZAOC |
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| 39. |
9.Find the value of cot 52°.cot 22". cot 38°.cot 68 |
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| 40. |
If x²+1\x²=38, then find the value of x³-1/x³ |
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| 41. |
Find the value of \frac{\cos 38}{\sin 52} |
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Answer» cos38 = cos(90-52) = sin52 cos38/sin52 = sin52/sin52 = 1 |
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| 42. |
Q. 13. Ifx-5.216,find x2 + |
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Answer» Like if you find it useful |
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| 43. |
43N]e 21४ 28 1308 || A ‘e ey TB By न |
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Answer» दिया गया है DE is parallel to BCAD/DB= AE/EC( थल्स प्री में से1.5/3= 1/AE15/30= 1/AEAE= 2cm |
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| 44. |
\int \frac { e ^ { 4 \operatorname { log } x } - e ^ { x \operatorname { log } 4 } } { e ^ { 2 \operatorname { log } x } - e ^ { x \operatorname { log } 2 } } d x |
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| 45. |
If & and B are the the zerose of the quadraticPolynomial food = 22 - PC ta then find the value ofis 2²+32 (it te |
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| 46. |
कक. 5हि के [0 २ कै व e and (H - o e4 ३८ 4 53 r\gfl.flb—dfi¢0_The J | of 2; ५५९ .- LI.“ 1 4 31:35 ष कक |
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Answer» side PR is the longest because the angle opposite to it is the largest among all. |
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| 47. |
The length of a verandah is 3m more than its breadth. The numerical value of its area is equal to the numerical value of its perimeter. Find the length and breadth of the verandah. |
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| 48. |
d the numerical value of x fromgram given below.75° |
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| 49. |
Q6. Find the diameter of a cylinder whose height is 5cm and numerical value of volume equal to numerical valeeof curved surface area. |
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Answer» Volume V = π * r ^ 2 * h Curved Surface Area A = 2 * π * r * h We know that V = A Therefore ,π * r ^ 2 * h = 2 * π * r * h Therefore r ^ 2 = 2 * r or r = 2Thus d=r×2=2×2=4 cm |
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| 50. |
. If 00 x 90°,state the numerical value of x for which sin x° = cos x。 |
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