Saved Bookmarks
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. |
Two cylinders of equal volume have their heights in ratio 1:2 find the ratio of their radius |
|
Answer» thnx |
|
| 2. |
Two cylinders of same volume have their radii in theratio /5:2. What will be the ratio of their heights? |
| Answer» | |
| 3. |
The diameters of two cylinders are in the ratio of 2:.3. Find the ratio of their heightsf their volumes are equal. |
| Answer» | |
| 4. |
5. Two cylinders of equal volume havetheir heights in the ratio 2:1. What is theratio of their radii?(D) 1:42 |
| Answer» | |
| 5. |
Find compound interest on12600 for 2 years at 10% per annum compounded annually. |
|
Answer» Show properly |
|
| 6. |
.Show that sin (180°+A), tan(360°-A) see (180-A)-tan"Acos 2(904A)cosec2 A. sin(180°-A) |
|
Answer» =tan^3A |
|
| 7. |
a Find C.I. on12600 for 2 years at 10% per annum compounded annually. |
|
Answer» thax |
|
| 8. |
14. The polynomial p(x) = x - 2x + 3x? - ax + b when divided by (x - 1) and(x + 1) leaves the remainders 5 and 19 respectively. Find the values of aand b. Hence, find the remainder when p(x) is divided by (x - 2). |
|
Answer» aata nhi h bhai kya karu |
|
| 9. |
a=34,b=56,c=67,hence is ax+bx+c=456.find the value x. |
|
Answer» ax + bx + c = 45634x + 56x + 67 = 45690x = 456-6790x = 389X = 4.32 |
|
| 10. |
1+ £2 = 180° + 4, सिद्ध कीजिए । |
| Answer» | |
| 11. |
2 180 30%tan’ 30° TR B |
| Answer» | |
| 12. |
30. What must be added to the polynomial f(x) x4+2x3 -2x2+x-1 so that the resultingpolynomial is exactly divisible by x2 + 2x-3? |
| Answer» | |
| 13. |
1. The simple interest on6250 at 4% per annum for 6 months is(c)al 125(b)1 5075 |
|
Answer» Interest = P*R*T / 100= (6250*4*0.5)/100= Rs. 125 |
|
| 14. |
11. The angles of a quadrilateral are in the ratio 1:2:34. The smallest and the larges(a) 150, 750(b) 180, 820(c) 180, 720(d) 159, 720Lullaloramic |
|
Answer» here ans is a (C) 18°,72° c aara he mere dosht plz yaer answer is c -18 and 72 all the options are incorrect. kudo is right 👌All the options are incorrect. see the solution below👇 c is the correct answer the correct answer is [C] 18°, 72° |
|
| 15. |
5. A certain sum of money becomestimes ofitself in 6 years at a certain rate. Find therate(a) 135(b) 10222(c) 117%(d) 10% |
|
Answer» money becomes 7/4 times so interest is 3P/43P/4=P*R*6/100so R=(100×3)/(4×6)=12.5% |
|
| 16. |
9. What must be added to polynomial fi)22x21 so that the resultingpolynomial is exactly divisible by 2 |
|
Answer» P(x) = x^4 + 2x^3 -2x² + x -1 d(x) = x² + 2x -3 We divide p(x) by d(x) & find the quotient q(x) & remainder r(x) P(x) = d(x) * q(x) + r(x) We find that r(x) = (-x +2) Hence, the additive inverse of remainder should be added to p(x), so that p( x) will be exactly divisible by d(x) Additive inverse of -x + 2 = (x-2) Ans |
|
| 17. |
d) What should be added to the polynomial x2-5x+4, so that (x - 3) is factothe resulting polynomial?2(ii) 4(iv) 5 |
|
Answer» Given polynomial x^2 - 5x + 4 If (x - 3) is factor of polynomial then for x = 3 given equation will be equal to 0 Put x= 3 in given equation of polynomial3*3 - 5*3 + 4 = 9 - 15 + 4= 13 - 15 = - 2 Therefore 2 should be added to given polynomial. thanks |
|
| 18. |
What must be subtracted from the polynomial plx)x203-13x2-12x +21 so thatthe resulting polynomial is exactly divisible by x2-4x+ 3? |
| Answer» | |
| 19. |
14. What must be subtracted from the polynomial p(x)2x3-13x2 -12x + 21 so thatthe resulting polynomial is exactly divisible by x2-4x 3? |
| Answer» | |
| 20. |
8. (i) If the ratio of the radii of two cylinders of equal heights is 2:3, then find the ratio of their cur(ii) The ratio of the height of two cylinders of same base radii is 3:2. Find the ratio of theirsurfaces.surfaces |
|
Answer» If you like the solution, Please give it a 👍 Fantastic solve |
|
| 21. |
9595 \div 19-57 \div 3=? |
| Answer» | |
| 22. |
13. What must be subtracted from the polynomial f(x)2r3-13x2-12x+21sothatthe resulting polynomial is exactly divisible by 2 -4x +3? |
| Answer» | |
| 23. |
(iii) CP =1540, loss = 4% \ |
| Answer» | |
| 24. |
FACTORISECa toj -49 |
|
Answer» 49=7^2hence(a+b)^2-7^2anda^2-b^2=(a-b)(a+b)hence(a+b-7)(a+b+7) |
|
| 25. |
3 ^ { 7 } \div 3 ^ { 9 } = 3 ^ { \square } |
|
Answer» 3^7-9 3^-2 3-²is the answer. |
|
| 26. |
Find the area of a rectangle whose(1) length =46 cm and breadth = 25 cm |
| Answer» | |
| 27. |
Arif took a loan of80000 from a bank. If the rate of interest is 10% per annum,find the10.difference in amounts he would be paying after 12 yearscompounded annually. (ii) compounded half-yearlyif the intereif the interest is. The difference between the compound interest and simple interest on a certain sum for2years at 75%) per annum is? 360. Find the sum. |
|
Answer» Amount Rs 88000+Rs 4400 this your mistake |
|
| 28. |
a farmer has borrow 2400 rupees at 12% interest per annum find the amount to pay at the end of two and half years |
|
Answer» si= p×r×t/100 2400×12×2.5 =864 amount = 864+ 2500 =3264 |
|
| 29. |
35 :Find the interest onthe amount to be paid at the end of 3 years.800 for a period of 3 years at the rate of 4% per annum. Also, fin |
| Answer» | |
| 30. |
The measure of each angle of an equilateral triangle is(2) 180°(3) 60°(4) 40° |
| Answer» | |
| 31. |
A sum of 2500 is borrowed at a rate of 12% per annum for 3 years. Find thesimple interest on this sum and also the amount to be paid at the end of 3 years. |
| Answer» | |
| 32. |
ice the shhaier angle.PQRS is a parallelogram where PQ is 12 cm and perimeter is 46 cm. Find the lengthof each side of the parallelogram. |
|
Answer» sides of parallelogram is 11,11,12,12 |
|
| 33. |
เคนเคฟ + 950 =1 64% - 900+ 7 |
|
Answer» Please hit the like button if this helped you out. |
|
| 34. |
8. Saurabh invests 48,000 for 7 years at 10%per annum compound interest. Calculate :(i) the interest for the first year.(ii) the amount at the end of the second year.(iii) the interest for the third year. |
|
Answer» Compounding is once at the end of the year.So the interest for n year is P (1+r/100)^n - P, where P = principle, r = rate of interest and n = number of years..Given P= 48000. r = 7 and n = 1. Substituting these values in the formula, we get:Interest for the first year = 48000(1+7/100)^1 - 48000 = 480*7 = 3360 rupees. |
|
| 35. |
1. Find the SP when:(i,CP-7 950, gain06%(iii) CP1540, loss-4% |
| Answer» | |
| 36. |
Anita takes a loan of Rs 5000 at 15% per year as rate of intrest. Find theinterest and the amount to be paid at the end of 3 years. |
|
Answer» interest 2250 &amount to be paid at end 7250 total entrest amount of the end of three years to 2100rs interst rate 2250+5000( amount )=7250 ruppess interest 2250& amt 7250 Interest 2250Total Payable amount 7250 total interest and amount paid by Anita Rs. 17250.00 the answer is Interest =PxRxT/100=5,000×15/100×3=2,250/- total amount to be paid is 5,000+2,250 = 7,250/- here P is principal R is Rate of Interest T is Time interest will be 2250and the amount is 7250 si=p×r×t/100=5000×15×3/100=2250 2250 is rate of interest in 3 years and total amount paid by ankita is 7250 the answer to this question is interest 4750rs and amount is 9750 interest 2250 and total amount 7250 intrest amount 2604 and end 7604 2250 correct ans. h 😁😁😁😁😁😁😁😁 according to simple interest the interest is 2250 rupees and the total amount to be paid is 7250 rupees intrest rate 2250+5000=7250 amount tu and in 2250 3 year interest tell me whether this amount on simple interest or compound interest 2250 of interest and amount is paid 7250 interest is amount 2250 7250 is the right answer interest=2250 and total amount at the end of 3 year=7250 2250 interest and 7250 paid at the end 2250 interest and 5000 at the end intereste 2250 & amount to be paid at 7250 interest 2250 amount to be paide 7250 2250 per year amount interest 2250 and amount 7250 P=5000R=15%T=3 Years S.I=P×R×T =5000×15×3/100 =2250Amount =P+I =5000+2250 =7250 2150 interest of 3 Year total amount 7150 5000×15÷100=750 interst per year 3 year interst 3×750=2250total amount 5000+2250 2250 is right answer interest= 2250Total=7250 7250 interest + Rupea 2250and end me amount 7250 ru interst=2250 and amount =7250 loan5000+2250interest=7250 right answer is 2250 1 year of interest 750 & of 3 year 2250 paid total 7250 2175 is interest.loan amount=5000total amount=5000+2175 =7175 15/100*5000 ans=₹2250 Ans is the 5985 15% of 5000 is 5985 Interest is 2250 and total amount is 7250 interest=2250+5000=7250 interest 2250 and amount to be paid end of 3 years 7250 intrest is 22250 and total amount is 7250 2250 after 3years intrest paid karegi interest amount 7250 the answer will be 7250 15/100=3/203/20*3=9/205000*9/20=22505000+2250=7250 I=2250,P=5000,A=7250 that's loan interest became a 2250$ & his paid a amount 7250$ amount tu and in 2250 3year interest the correct answer is 750 at the end of 3rd year interest to be paid is rupees 2250 75% amount is 8750 rupees interest 2250 amount 7250 2250rs.................. दर 2250 मुलधन 5000 कुल 2250+5000=7250 2250 is the Interest.SI=Principal+Interest2250+5000 =7250 ye Raju mera chasma la re jara comment padh lu to😂😂😂 interest will be 2250 and amount to be 7250 2250 interest nd paid after 3 year P = 5000, R= 15% , T = 3yearsS.I = P×R×T/100 = 5000×15×3/100 = 2250₹Amount = S.I + P = 2250+ 5000 = 7250₹ total payable amount 7250 total interest is 2250 and he paid 7250 |
|
| 37. |
1. Find the SP when:(i), CP-7 950, gain-6%O7o(iii) CP1540, loss4% |
|
Answer» (i) Gain percentage = ((SP - CP)/CP)*100 6 = ((SP-950)/950)*100 5700/100 = SP - 950 SP = 900 + 57 = 957Rs(ii) Loss percentage = ((CP - SP)/CP)*100 4 = (1540 - SP)/1540)*1009240/100 = 1540 - SP SP = 1540 - 92.4 = 1447.6Rs |
|
| 38. |
7'''' न 18 2९ ४2 100 1R T =950 24k ()(PO AT |
|
Answer» cosθ = √3/2 So, θ = 30° |
|
| 39. |
The population of a town is 10,000. It increases by 10% during the firstyear. During the second year, it decreases by 20% and increased by30% during the third year. What is the population after 3 years? |
|
Answer» Given,Starting population = 10000 Population after first year= 10000(1 + 10/100)= 10000(11/10)= 11000 Population after second year= 11000(1 - 20/100)= 11000(4/5)= 2200*4= 8800 Population after third year= 8800(1 + 30/100)= 8800(13/10)= 880*13= 11440 Therefore, Population after 3 years = 11440 |
|
| 40. |
find the compound intrest on rupees 12,000 at rate of intrest of 12% for one year , if the intrest is being compounded half yearly |
|
Answer» P= 12000R=12℅T=1 On doing compounded half yearly R=12÷2=6T=1×2=2 Amount= P×{1+R÷100}*T 12000×{1+6÷100}×2 =13483.2CI=Amount -Op 13483.2-120001483.2 |
|
| 41. |
In Regeniawhicho suum2.205 n12 |
| Answer» | |
| 42. |
find the intrest paid by a Barrow after a period of 3 years on a loan of rs 2000 at 10% p.a, intrest being compounded annuly |
| Answer» | |
| 43. |
ULOU.10. On a certain sum of money, lent out at CI.,interests for first second and third years are1,500; 1,725 and 2,070 respectively.Find the rate of interest for the (i) second year(ii) third year. |
|
Answer» The compound interest accumulation formula is given by : A = P(1 + i)^n For the first year : P(1 + i) = 1500 P = (1500)/(1 + i) Second year : P(1 + i)² = 1725 P = 1725/(1 + i)² We have that the principle in the first year is equal to the principle in the second and the third year. Equating the two : 1725/(1 + i)² = 1500/(1 + i) 1725/1500 + (1 + i)²/(1 + i) 1.15 = 1 + i i = 1.15 - 1 i = 0.15 i = 15% Interest rate is thus 15%. |
|
| 44. |
On a certain sum the interest paid after 3 years is Rs 450 at 5% rate of interest perannum. Find the sum.2. |
| Answer» | |
| 45. |
a certain sum the interest paid aner 3 years it 450 at 5% rate of interestum. Find the sum. |
| Answer» | |
| 46. |
on a certain sum the interest paid after 3yrs is rs 450 at 5% rate of interest per annum find the sum |
| Answer» | |
| 47. |
Area =Are your guesses close enough to the exact calculated values? If yes, give a pattoyourself and share it with your parents. Celebrate it by playing Ludo, Chess or a gameof your own choice with your parents.Kouledge Check 15.1 --Find the area of a rectangle whose(1) length = 36 cm, breadth = 15 cm(in) length = 24.5 cm, breadth = 5.8 cm(ii) length = 22.8 cm, breadth = 7.5 cm(iv) length = 7.9 m, breadth = 90 cm.New Learnwell (ICSE) Mathema |
|
Answer» area of the rectangle = length × breadth 1) 36×15=540cm^22)22.8×7.5=171cm^23)24.5×5.8=142.1cm^24)7.9×90=711cm^2 Length =36 cmBreath =15The situation of rectangle is =2 multiple [36+15]36 +15=51: we have to multiple 51 by 2=102 cm 102 cm is the answer 102cm is the correct answer |
|
| 48. |
me ve the room temperature 8 hours after the process begins?12. In a test, there are 100 questions. +3 marks are given for each correct answer, -1 mark for eachincorrect answer and 0 marks are given for not attempting the question. Rakesh scores 190 marks in atest by answering 70 questions correctly. Find the maximum marks in the test and also find the numberof incorrect answers given by Rakesh. |
|
Answer» thanks |
|
| 49. |
s.p=₹720.80 and gain=6% |
|
Answer» gain = 6100 Perce= 720.801 percentage= 7.20106 per= 7.20*106 gain =6100perce=720.801 percentage =7.20106 per=7.20*106 106 Per=720.80100 per=XX=100*720.80/106=680cost price=680gain=720.8-680=40.8 |
|
| 50. |
(1) CP=950. gain6% |
|
Answer» CP = 950Gain = 6% let SP be So, according to given question, SP = 950 + 6% of 950 950 + 6/100 × 950 950 + 57 1007 So, SP = ₹ 1007 |
|