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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. |
Find CI on a sum ofă8000 for2 years at 5% per annumcompounded annually1.( h |
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| 2. |
-CP-7 950, gain = 6% |
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Answer» Gain%=(SP-CP/CP)*1006/100=SP-950/95057=SP-950SP=1007Rs |
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| 3. |
i) CP = Rs 950, gain = 6% |
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Answer» C.P = 950 and gain = 6% of C.P = 6*950/100 = 57 so, S.P = C.P + gain = 950+57 = 1007 thanks |
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| 4. |
The compound interest on a certain sum of moneyat a certain rate per annum for two years is Rs. 2050and the simple interest on the same amount ofmoney at the same rate for 3 years is Rs. 3000. Thenthe sum of money is |
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| 5. |
A sum of money is lent at simpleIntrest whicho amount to Rs 3192in 3 years and to RS 37201in 50 years. Find the suniOF money and the rateof Intrestu od |
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Answer» the correct answer is 11% |
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| 6. |
/200 at 12.75% per annum for 9 months.A certain sum of money amounts to 1560 in 2 years and Ĺş 2100 in 5 years. Find therate percentage per annum. |
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Answer» suppose the amount is xthenAmount = SI + principal X + 2rx = 1560......................iX + 5rx =2100 .....................iiSolve the equationx=1200 and r is 15r= rate of interestx = amount2rx and 5rx are used |
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| 7. |
d the SP when:CP =950, gain = 6%Il |
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| 8. |
EMATICS-a |
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| 9. |
. Find SP when: 15,20(a) CP= 5920 and gain = 6% |
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Answer» S.P= cp+gainS.P=920+6%of 920S.P=920+6/100 of 920S.P=920+6×9.2S.P=920+55.2S.P=975.2 Given cost price of the book =920Profit%=6%selling price = (100 + profit%)cost price/100 = (100 + 6%) × 920/100 =(106)× 92/10 =(106×92)/10 =9752/10 =975.2 Therefore, the selling price of the book is 975.2 = Ans✓
S.P =975.2 is the best answer S.P.=975.2 is correct answer. s.p=cp+gain. s.p=920+6%of920. s.p=920+6/100of920 s.p=920+55.2. s.p=975.2 s.p = c.p + gains.p= 920 +6% of 920s.p= 920 +6/100 of 920s.p = 920 + 6×9.2s.p = 920 + 55.2s.p = 975.2 S.P = 975.2 is the correct answer sp = cp*(100+profit%)/100sp= 900*(100+6)/100sp= 900*106/100sp= 95400/100sp=954 answer SP=975.2 rs is correct answer S. P=cp+gainS. P=920+6%of 920S.P=920+6/100 of 920S.P=920+6×9.2S. P=920+55.2S.P=975.2 answer= s.p=975.2 is correct Selling Price= 975.2 Cp=R 950 gain =6%6% of 950 rs +rs 950=sp 950*6)/100.gain rs 57 CP=r950 gain =6%6%of 950 rs 950 =so <950*6>/100. gain rs 57 the correct answer is the55.2 CP=r 950 gain =6%6% of 950 rs 950 = so <950*6>/100. gain rs 57 sp=975.2 is right answer answer = s.p 975.2 is thi best ans Gain = 6 %of ₹920 = 6/100 × 920 = ₹ 552/10 =55.2Sp = Gain + cp = 55.2+920 = 975.2 Formula of SP=CP-GainSP=920+6%÷100×920SP=920+6×9.2SP=920+55.2SP=975.2 ANSWER=975.2 sp=cp+gain. sp=920+6%of 920sp=920+6/100of920sp=920+55.2sp=975.2 S.P=975.2 ,your answer is here s.p 975.2 exect answer correct answer is s.p =975.2 Sp= 100+gain%×cp 100100+6%×920100=5520 Answer of the question is s.p= 975.2 S.P=C.P+GainS.P=920+6/100 of 920S.P=920+6×9.2S.P=920+55.2S.P=975.2 SP=(CP)×(100+profit percent)/100 cost price = 920 rspercentage Gain = 6%we know,Selling Price = Cost price * percentage profit /100soSP = 920 * 106/100SP = 975.2 rs 975.2is right answer s.p=cp+gains.p=920+6%of920s.p=920+6/100of920s.p=920+6×9.2s.p=920+55.2s.p=975.2 s. p= 975.2 is the best answer for questions S.P.=C.P+ GainGAIN=6%of920S.P.=920+55.2S.P.=975.20 975.2 is the best answer of this question 975.2 is gain cp=920 and sp=975.2 selling price = (100+profit%)cost price/100 = (100+6%)×920/100=(106)×92/10=(106×92)/10=9752/10=975.2 gain =cp-spsp=cp+gainsp=975.20+6/100*975.20 =975.20+58.512 =1033.712 975.2 is the correct answer cp = 920 then 6% of 100 is 55.2 so sp= 920 + 55.2 = 975.2 it's answer is 975.2 s.p =975.2 isthe best answer s.p=cp+gains.p=920+6%of920s.p=920+6/100of920s.p=920+6×9.2s.p=920+55.2s.p=975.2 CP = ₹ 920gain =6%SP = ? SP = CP × 100 ÷ 100 + gain%SP = 920 ×100 ÷ 100 + 6SP = 92000 ÷ 106SP = given,CP=920Gain=6%of the CP =6%of 920=6/100 × 920=55.2gain=55.2 SP=CP+gain SP=920+55.2=975.2 SP=975.2 |
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| 10. |
4. The average of marks in Math-ematics for 5 students was foundto be 50. Later, it was discoveredthat in the case of one studentthe marks 48 were misread as84. The correct average is: |
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Answer» Average of marks = sum of marks/ no. of students Then,50 = sum of marks/5sum of marks = 250 Difference between actual marks and misread marks= 48 - 84= - 36 Then,Sum of marks = 250 - 36 = 214 Correct average = 214/5 = 42.8 |
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| 11. |
Find С.Р. when (i) S.P. = Rs 530, gain = 6%, |
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Answer» we have formulaS.P.={(100+gain%)C.P.}/100530=((100+6)/100)×C.P.530=106/100 ×C.P.C.P.=530×100/106C.P.=Rs500 |
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| 12. |
an ald mode of these scoreS.2.-In a mathcorded:ematics test given to 15 students, the following marks (out of 100) are41, 39, 48, 52, 46, 62, 540, 96, 52, 98, 40, 42, 52, 60Find the mean, median and mode of this data. |
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| 13. |
1. Find the SP when:(i) CP-7950, gain6% |
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Answer» CP = ₹950 Gain% = 6% So, Gain amount = 6% of ₹ 950 = 6/100 × 950 = ₹57 So, SP = CP + Gain = 950 + 57 = ₹ 1007 Hit like and share |
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| 14. |
the figureoken at the pointand let CB take theThen. ACO m. AD 8 m and CD) - CBth - 6) m.From right ADAC by Pythagoras' theorem. we have:CD = AC + AD(- 6) = 6° +82= 36 +64 = 100 = (10)= (h- 6) = 10 = h = 1046) - 16 m.Hence, the original height of the tree was 16 m.Find the length of the hypotenuse of aEXERCISE 150chuse of a right triangle, the other two sides of which measure9 cm and 12 cm.The hypotenuse of a right triangle is 26 cm long. If one of the remaining two sideslong, find the length of the other side.The length of one side of a right triangle is 4.5 cm and the length of its hypotenuse isFind the length of its third side.length of its hypotenuse is 7.5 cm.Hist. Let the third side be x cm. Then,x =7.5) -(4.5)' = 17.5+4.5)(7.5-4.5) = (12x 3) = 36 = 16).The two legs of a right triangle are equal and the square of its hypotenuse is 50. Find thelength of each leg.5. The sides of a triangle measure 15 cm, 36 cm and 39 cm. Show that it is a right-angledtriangle.Hint. (39)-(36)" = (39+36)(39-36) = (753) = (5x5x3x 3) = (5x3)2 = (15)In right AABC, the lengths of its legs are given as a = 6 cm and b = 4.5 cm. Find the lengthits hypotenuse.The lengths of the sides of some triangles are given below. Which of them are right-angled(i) a = 15 cm, b = 20 cm and c = 25 cm-9cm h = 12 cm and c = 16 cm |
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Answer» ( hypotenuse)²= (side)²+(side)² hypotenuse=(9)²+(12)² =81+144 =225 =√225 hypotenuse=|15is the best answer (2). hypo=10²+side² 26²=10²+side² side²=676-100 side²=576 side=24is the best answer |
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| 15. |
The first term of a G.P. is 27 and its 8th term is 1/81 . Find the sum of first seven terms ofthe G.P. |
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| 16. |
4. The volume of a wooden block is 1440 cm^3. If its length and breadth are respectively36 cm and 8 cm, what is its height? |
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| 17. |
7. A certain sum of money yields Rs. 927 as compound interest in 2 years at 6% pa, compoundedyearly. Find the sum.(Rs. 7500) |
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| 18. |
10240 in 2 years at 6 2% per annum compounded annuallysum of money amounts toFind the sum.3 |
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| 19. |
L003 3 monthsyears.4Abhay borrowed? 1 6000 at 7 스% per annum simple interest. On the same day, he lent it toGurmeet at the same rate but compounded annually. What does he gain at the end of2 years?2 |
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| 20. |
A certain sum of money amounts to 5292 in two years and 5556.60 in three years, interest being compounded annually. Find rate percentage |
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| 21. |
सानित 530 3, समातरनाकु यतुष्ठीशना सामस। मेन, BRI समान 9, |
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Answer» એસી અને બીડી પેરેલલોગ્રામના ત્રિકોણ છે. સમાંતર ચક્રના પ્રત્યેક ત્રાંસા સમાંતર ત્રિકોણને બે સમાંતર ત્રિકોણોમાં વહેંચે છે. ΔADC ≅ Δ એબીસી con અભિનંદન ત્રિકોણની સુસંગત બાજુ સમાન છે. ⇒ સીડી = એબી ⇒ એડી = બેચન્સ સમાંતર ચક્રની વિરુદ્ધ બાજુઓ સમાન છે. Consider a parallelogram ABCD.AC and BD are the diagonals of the parallelogram.Each diagonal of the parallelogram divides the parallelogram into two congruent triangles.ΔADC ≅ ΔABC∴ Corresponding sides of congrunt triangles are equal.⇒CD = AB⇒ AD = BCHence the opposite sides of the parallelogram are equal. |
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| 22. |
Wematics for Class 8A sum of money amounts to 10240 in 2 years ao per annum, compounded annually.Find the sum. |
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| 23. |
3. यदि 25 टनं किसी न के 24 दिन में आते हैं - दनकितने टर्म इस इन द 20 दिन में कर स्को?(a) 20(530(d) 2544 |
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Answer» Total work is 25 × 24 = 600Man required to complete the work in 20 days = 600 ÷ 20 = 30 days |
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| 24. |
_sec37_का मान ज्ञात करेंcosesc 530 |
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| 25. |
Mathematice for Clase16. A sim of money amounts to t 10240 in 2 years at 62% per anmim. compounded anFTnd the sun. |
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Answer» Please hit the like button if this helped you |
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| 26. |
Find C.P. when (i) S.P' Rs 530, gain6%, |
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Answer» SP = (100 + Gain%)/100 * C.P. 530 = 100 + 6/100 * C.P. 924 = 106/100 * C.P C.P = 924 * 100/106 C.P = 871. 69. |
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| 27. |
Show that "the range of the first seven mutliples of the smnme number is also a multiple of that prime number" |
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Answer» 1 2 |
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| 28. |
EXERC1. Find the area of a rectangle whose:(i) length 36 cm, breadth 15 cm |
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| 29. |
2 tan2 60Find the value of 1-tan' 30+ Cosec"450-3 cot245°+8in2300, |
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Answer» Hit like if you find it useful |
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| 30. |
tan 45°+0-1-tan θHil(.ld satan|450 +6H1 |
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| 31. |
Thus, 25% of 4.46 kg = 1.115EXERC1. Find :(i) 20% of 60(iv) 125% of 10(vii) 25% of 100(v)(viii) |
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Answer» 20% of 60 = (20/100) × 60 = 12125% of 10 = (125/100) × 10 = 12.525% of 100 = (25/100) × 100 = 25 so here20 percent of 60 20% of 60 = (20/100) × 60 = 12125 percent will be 125% of 10 = (125/100) × 10 = 12.5 25 percent will be25% of 100 = (25/100) × 100 = 25 |
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| 32. |
the number line and represent the following rational numbers on it:5-7(iv)1V4 |
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| 33. |
लपा पर एसए32.सिद्ध कीजिएsin 8-cos 9 sin -cos esine+cose sine-cose 2sin20-1अथवाScos260°+4c052300-tan-450का मान ज्ञात कीजिए।cos2600+sin2300 |२१आकति में ARC एक वत्त जिसकी त्रिज्या 14 सेमी है, का चतुर्थांश है तथा BC को व्यास मानकर एक अ |
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Answer» where's your school 2/2sin²0-1is the correct answer |
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| 34. |
2.7.If the system is in equilibrium(cos 530 3/5), then the value of 'P' is6N53TON |
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Answer» equilibrium means sum of the all resultant force acting on it is equal to 0 |
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| 35. |
dia for 3 years 8 months at 12% perShanta borrowed 6000 from the State Bank of Inannum. What amount will clear off her debt?2. |
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| 36. |
Three circles of unit radii are inscribed inan equilateral triangle. Find the area of thetriangle.で币研阿底僦 可ず37adf 骊传俪ァ1丽77JT(A) (12+ 8 3 )square unit(B) (6 + 4/3)square unit(C) (8 + 42) square unit(D) (12 6/2) square unit |
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Answer» option b, because after calculating all the options, that resulted in 12.93 |
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| 37. |
The Sum of First Four Prime Number is? |
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Answer» First four prime numbers are2,3,5,7Their sum5+5+7 = 17 |
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| 38. |
2π兀6sin cosuation: |
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Answer» sin2A = 2sinA cosAcos2A = 2cos²A - 1 = 1 - 2sin²A∫ sinx dx = -cosx + C∫ cosx dx = sinx + C ∫ sin⁴x cos⁶x dx= ∫ sin⁴x cos⁴x cos²x dx= (1/16) ∫ 16 sin⁴x cos⁴x cos²x dx= (1/16) ∫ (2 sinx cosx)⁴ [ (cos2x + 1)/2] dx= (1/32) ∫ (2 sinx cosx)⁴ (cos2x + 1) dx= (1/32) ∫ (sin2x)⁴ (cos2x + 1) dx= (1/32) ∫ (sin⁴2x cos2x + sin⁴2x) dx= (1/32) ∫ sin⁴2x cos2x dx + (1/32) ∫ sin⁴2x dx= (1/32) ∫ sin⁴2x cos2x dx + (1/32) ∫ (sin²2x)² dx= (1/32) ∫ sin⁴2x cos2x dx + (1/32) ∫ [ (1 - cos4x)/2 ]² dx= (1/32) ∫ sin⁴2x cos2x dx + (1/32) ∫ (1 - cos4x)²/4 dx= (1/32) ∫ sin⁴2x cos2x dx + (1/128) ∫ (1 - cos4x)² dx= (1/32) ∫ sin⁴2x cos2x dx + (1/128) ∫ (cos²4x - 2cos4x + 1) dx= (1/32) ∫ sin⁴2x cos2x dx + (1/128) ∫ cos²4x dx - (1/64) ∫ cos4x dx + (1/128) ∫ dx= (1/32) ∫ sin⁴2x cos2x dx + (1/128) ∫ [(cos8x + 1)/2] dx - (1/64) ∫ cos4x dx + (1/128) ∫ dx= (1/32) ∫ sin⁴2x cos2x dx + (1/256) ∫ (cos8x + 1) dx - (1/64) ∫ cos4x dx + (1/128) ∫ dx= (1/32) ∫ sin⁴2x cos2x dx + (1/256) ∫ cos8x dx + (1/256) ∫ dx - (1/64) ∫ cos4x dx + (1/128) ∫ dx= (1/32) ∫ sin⁴2x cos2x dx + (1/256) ∫ cos8x dx - (1/64) ∫ cos4x dx + (3/256) ∫ dx the only problem here is to solve the first integral in the above expression∫ sin⁴2x cos2x dxlet u = sin2x ==> du/dx = 2cos2x ==> dx = du/(2cos2x) ∫ sin⁴2x cos2x dx= ∫ u⁴ cos2x [du/(2cos2x)]= (1/2) ∫ u⁴ du= (1/2)(1/5) u⁵ + C= (1/10) u⁵ + C= (1/10) sin⁵2x + C The finishing step:∫ sin⁴x cos⁶x dx= (1/32) ∫ sin⁴2x cos2x dx + (1/256) ∫ cos8x dx - (1/64) ∫ cos4x dx + (3/256) ∫ dx= (1/32) (1/10) (sin⁵2x) + (1/256) (1/8)(sin8x) - (1/64) (1/4)(sin4x) + (3/256) x + C= (1/320) sin⁵2x + (1/2048) sin8x - (1/256) sin4x + (3/256) x + C ⇦ ANSWER |
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| 39. |
Exerc1. Simplify: (i) (25)° (33)? ; (i(29) * (32)3 |
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Answer» 1is the correct answer 1 is the best answer me the answer is 1(one) 1 is correct answer❤❤❤ |
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| 40. |
EXERCProve that:1. sincos2 |
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| 41. |
14). lFind the nature of roots for the quadratic cq uation 2x'- 5x+1-0. |
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Answer» Firstly find b²-4ac to find the nature of roots.b=√5, a=2, c=1 b²-4ac=(√5)²-4(2)(1)=5-8=-3<0So the roots are imaginary |
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| 42. |
What is the value of z in the figure shown below?4.369°434.3 |
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Answer» this figure is parallelogram..the sum of adjacent angle of parallelogram is 180°69+45+x=180°114°+x=180°x=180°-114°=66° |
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| 43. |
uation that can be used to find the missing length for each triangle:(iv)1V13107 |
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| 44. |
12 Shanta borrowed6000 from the State Bank of India for 3 years 8 months at 12%annum. What amount will clear off her debt? |
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| 45. |
(d) Shanta borrowed6000 from the State Bank of India for 3 years 8 months at12% per annum. what amount will clear off her debt? |
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Answer» Here , In this question,Principal = Rs 6000Time = 3 yr 8 months = 3 + 8/12 = 11/3 years.Rate = 12% per annumSo, Simple Interest = PRT/100 = (6000*12*11/3 )/100i.e. 2640.Now, Amount = Principal + Interest i.e 6000+2640 = Rs 8640.Rs 8640 is that amount which will clear off her debt. |
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| 46. |
2. Shanta borrowed6000 from the State Bank of India for 3 years 8 months at 12% perannum. What amount will clear off her debt? |
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Answer» thanks |
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| 47. |
Finndthe valueof\tan 9^{\circ}-\tan 27^{\circ}-\tan 43^{\circ}+\tan 3 |
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| 48. |
20111 x 21 x 11 |
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Answer» 20111*21*1120111*2314645641 |
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| 49. |
108 - 21 x + x ^ { 2 } |
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| 50. |
x ^ { 2 } - 21 x + 108 |
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Answer» x^2-21x+108 =x^2-12x-9x+108 =x(x-12)-9(x-12) =(x-9)(x-12) |
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