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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. |
b) Neha also joined the same painting class as Karishma. She made 58paintings in the class and 9 more at home. How many paintings didNeha make in all ?то |
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Answer» Neha has made sixty seven paintings 67 is the answer of the following |
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| 2. |
/ 57.TA, TB are tangents to a circle with centre o. ChordAB intersects TO at C.Giventofind AB1.8cm2. 12cm3.9cm4.4cm |
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Answer» 1/(OA)² + 1/(TA)² = 1/36 Multiply both sides by (OA)² (TA)²(TA)² + (OA)² = (OA)² (TA)² / 36 Since TA is a tangent of the circle, then TA is perpendicular to OA, so △OAT has a right angle at A. By Pythagorean theorem:(TA)² + (OA)² = (OT)² Therefore:(OT)² = (OA)² (TA)² / 36OT = OA * TA / 6 AB is perpendicular to OT. Therefore, AC is perpendicular to OTSince △OAT is a right triangle, dropping perpendicular from right angle A to side OT (at C) creates 2 right triangles that are both similar to △OAT △OAT ~ △OCA ~ △ACT By similar triangles:AC/OA = TA/OTAC = OA * TA / OTAC = OA * TA / (OA * TA / 6)AC = 6 AB = 2 * ACAB = 12 Option 2) 12 cm is correct. |
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| 3. |
an2ab(c), find the value of r-abIf x=a+h, |
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| 4. |
If AB and CD are two chords of a circle with centre O,such that C,O,D are collinear and AB=one third CD.If AB=3 cm, find radius of circle |
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| 5. |
dTatb 37Find the values of a, b, c and d if15+c ab | |
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Answer» After comparinga+b= 6d= 35+c= -1= c= -6ab= 8nowa+8/a= 6a^2-6a+8= 0a^2-2a-4a+8=0a(a-2)-4(a-2)=0a= 4,2b= 2,4 |
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| 6. |
1) Neha and Sunil have some apples. Neha says to Sunil, "If you giveme 12 of your apples, I will have twice the number of apples left!with you". Sunil replies," if you give me 12 of your apples I willhave the same number of apples left with you". Find the numberof apples with Neha and Sunil separately.Odded to than |
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Answer» Answer: Neha:84 Sunil:60 Step-by-step explanation:Let neha has 'x' apples and sunil has 'y' apples If sunil gives 12 apples to neha Neha is left with x+12 apples and sunil is left with y-12 applesNow neha has double the number of apples there with sunilSo x+12 = 2(y-12) x = 2y-36 .......(I) If neha gives 12 apples to sunilNow neha has x-12 apples and sunil has y+12 apples Now they both have same number of applesSo x-12 = y+12 x = y+24 ........ (II)Solving I and II 2y-36 = y+24 y =60Since x = y+24 x = 60+24 x = 84 So neha has 84 apples and sunil has 60 apples |
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| 7. |
21. A boy is cycling such that the wheels of the care making 140 revolutions per minute. Ifdiameter of the wheel is 60 cm, calculate the spper hour with which the boy is cycling. |
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| 8. |
find the least numberof s diget is exactlydivisble by 12,18,2130 |
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Answer» LCM (12, 18, 21, 30)= 1260 Therefore, least 5 digit number exactly divisible by 1260 is 12600 Lf x5 +51divisble by x+1remaind |
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| 9. |
(12) The sum of the digits of a two-digitnumber is 11. If the tens digit is 5 lessthan the ones digit, find the number. |
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| 10. |
P F, L " % % STe %))4(%’ >>< /L‘c'x(-ci छत,2 ० |
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| 11. |
The tens digit of a 2-digit number is less than the ones digit by 3. If the digits are reversed, thedifference between the two numbers is 27. If the sum of the digits is 11, find the numbers. |
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Answer» LetXXbe the1s1sdigit of a number,X−3X−3is the10s10sdigit, The number is10(X−3)+X10(X−3)+X. Reverse the number soX−3X−3is1s1sdigit andXXis10s10sdigits, Number becomes10(X)+(X−3)10(X)+(X−3) The SUM of digits is1111hence X+(X−3)=11X+(X−3)=11 2X=142X=14 X=7X=7 10s10sdigit=4digit=4,1s1sdigit=7digit=7,the number is 47. The reverse is7474, and74−47=2774−47=27istrue. |
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| 12. |
75×8×2×25 |
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Answer» 75x8x2x25=75×8×50=75×400=30,000 30,000 is the right answer 30,000 this is the correct answer answer is thirty thoushand only 30,000 is the correct answer of the given question. 30000is correct answer 75×8×2×25=75×8×50=75×400=30000 30000 is the right ans 30000is correct answer correct answer is 30000 the correct answer is 30,000 of the following question 75×8×2×25=30,000 is the answer of the following question 30,000 is the right answer 30,000 is the right answer 30000 this is the answer is yes sir I am his only son of a great day and I 30000 is the correct answer |
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| 13. |
\operatorname { log } \frac { 75 } { 16 } - 2 \operatorname { log } \frac { 5 } { 9 } + \operatorname { log } \frac { 32 } { 243 } |
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| 14. |
\log \left(\frac{75}{16}\right)-2 \log \left(\frac{5}{9}\right)+\log 32-\log 243-\log 2 |
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Answer» Given log(75/16) - 2log(5/9) + log(32/243). ------ (1) We know that a log(b) = log b^a. 2 log(5/9) = log (5/9)^2. We know that log(a) - log(b) = log(a/b). log(75/16) - log(5/9)^2 = log(75/16/(5/9)^2. = log (75/400/81) = log (75 * 31/400) = log (6075/400) = log (243/16) ----- (2) Substitute (2) in (1), we get = log (243/16) + log(32/243) We know that log a + log b = log ab. log(243/16) + log(32/243) = log(243/16 * log 32/243) = log(243 * 32/16 * 243) = log(32/16) = log 2. but answer is 0 |
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| 15. |
6. If the zeros of the polynomial f(x)-122 +39x+k are in A.P., find the valueof k |
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| 16. |
Find the ones digit from the cube of each of the following numbers.a. 2205 S2.b. 1149 |
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Answer» a. cube of 2205 will have 5 in ones place as 5 is multiplied three times b. cube of 1149 will have 9 in ones place as 9 is multiplied three times |
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| 17. |
14. A boy is cycling such that the wheels of the cycle are making 140 revolutioper minute. If the diameter of the wheel is 60cm, calculate the speed per hwith which the boy is cycling. /O |
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| 18. |
HUMIDE UN THUNon of 10 suchThere are ten lakh people in a district. What would be the pdistricts? / |
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Answer» 1000000*10=10000000(1crore) 100000000 (1 crore) is the answer of the following 1000000 x 10 = 10000000(1 crore is the answer) 1 district =100000010 district =10*1000000=10000000 1 crore is the correct answer of the given question 1 District =100000010 District =10*1000000 =10000000 1000000 x 10= 10000000 1 crore is the answer 10,00,000 People in a district × 10 district = 1 crore population in the 10 such district. |
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| 19. |
9. If 3A-4D10. İf A : B 3:4, B:C-5:7, find A: B : C. |
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Answer» A:B = 15:20 B:C = 20:28 A:B:C = 15:20:28 |
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| 20. |
(b) Given A=[-.].B-Bc-[:]and D-[:],3 -2B111, Cfind AB + 2C -4D |
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| 21. |
1§ 21% 8४0 ॥४४ WL =4d WO 7 =Dd bk ‘hisjieशाह कद 1% 10७ Debie ‘b 6171 Bjdble- |
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Answer» हल [प्रश्न हल करने की योजना(a) सबसे पहले आकृति में दिये गये अर्धवृत्त का क्षेत्रफल ज्ञात करें। (b) त्रिभुज PQR का क्षेत्रफल ज्ञात करें (c) अर्धवृत्त के क्षेत्रफल में से त्रिभुज PQR का क्षेत्रफल घटाने पर छायांकित भाग का क्षेत्रफल ज्ञात हो जायेगा।] दिया गया है, PQ = 24 cm PR = 7 cm अत: छायांकित भाग का क्षेत्रफल = ? हम जानते हैं किअर्धवृत्त में बना हुआ कोण एक समकोण होता है।. यहाँ चूँकि कोण RPQ अर्धवृत्त के अंदर बन रहा है, अत: यह एक समकोण है। अर्थात ∠ RPQ = 900 अब (त्रिभुज) Δ RPQ में पाइथागोरस प्रमेय के आधार पर QR2= PQ2+ RP2 = (24 cm)2+ (7 cm)2 = 576 cm2+ 49 cm2 = 625 cm2 अब चूँकि QR = दिये वृत्त का व्यास = 25 cm अत: त्रिज्या, r = 25/2 = 12.5 cm अर्धवृत्त के क्षेत्रफल की गणना हम जानते हैं कि अर्धवृत्त का क्षेत्रफल = 1/2 ×π r2 अत:, दिये गये अर्धवृत्त का क्षेत्रफल = 1/2 × π (12.5 cm)2 अत: दिये गये अर्धवृत्त का क्षेत्रफल = 245.53 cm2 त्रभुज के क्षेत्रफल की गणना त्रिभुज, PQR, में आधार = PR = 7 cm और ऊँचाई PQ = 24 cm हम जानते हैं कि त्रिभुज का क्षेत्रफल = 1/2 × ऊँचाई × आधार अत: (त्रिभुज) Δ PQR का क्षेत्रफल = 1/2 × 24 cm × 7 cm = 12 cm × 7 cm = 84 cm2 अत: Δ PQR का क्षेत्रफल = 84 cm2 अब दिये गये आकृत्ति में छायांकित क्षेत्र का क्षेत्रफल =अर्धवृत का क्षेत्रफल – त्रिभुज PQR का क्षेत्रफल = 245.53 cm2– 84 cm2 = 161.53 cm2 अत: प्रश्न में दिये गये छायांकित क्षेत्र का क्षेत्रफल = 161.53 cm2 hit like if you find it useful thanqu |
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| 22. |
कि. i x&’ ‘ ६ c=4d+x (O |
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Answer» x+y=5-------(1)2x-3y=4----(2)multiply eqⁿ1 by 33x+3y=15-----(3)(1)+(3)5x=19x=19/5pur in eqⁿ119/5 + y=5y=5-19/5y=6/5 |
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| 23. |
) V3 cos 23°—sin 23°=2 |
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Answer» If you like the solution, Please give it a 👍 |
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| 24. |
1.Find the areas of the rectan(a) 3 cm and 4 cm |
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Answer» Area = length × breadth = 3cm × 4cm = 12 cm^2 |
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| 25. |
gleThe perimeterof a rectangle is 72 cm. The length is 3 times the width. Find the area of the rectan |
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Answer» sorry foe inconvenience as in previous solution that length is 2 times of x |
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| 26. |
28) A square paper is folded into halfbreadthwise. The perimeter of the rectanformed is 12 cm. What is the area of thesquare? |
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| 27. |
Long Answer Questions:wire is in the shape of a square of side 10 cm. If the wire is rebent into a rectangle oflefind its breadth. Which encloses more area, the square or the rectanctangle of length 12 |
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| 28. |
(a) 8(b) 12(c) 24(d) 3510 The H.C.E of two numbers is 8. Which one of the following can never be their LCM2(a) 24(b) 48(c) 56(d) 60 |
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Answer» hcf * lcm = that number so HCF of two numbers is 8.This means 8 is a factor common to both the numbers.LCM is common multiple for the two numbers.So, the required answer is 60, |
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| 29. |
\left. \begin array l \operatorname cos ^ 2 30 ^ \circ \operatorname cos ^ 2 45 ^ \circ %2B 4 \operatorname sec ^ 2 60 ^ \circ %2B \frac 1 2 \operatorname cos ^ 2 90 ^ \circ - 2 \operatorname tan ^ 2 60 ^ \circ = ? \\ \text (a) \frac 73 8 \quad \text (b) \frac 75 8 \end array \right. |
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Answer» Option-Dis rightthanks cos30= √3/2cos45= 1/√2cos60= 1/2sin30= 1/2sin60= √3/2tan 60= √3 so3/4*1/2+4(4)+1/2(0)-2(3) 3/8+16-63/8+1083/8 |
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| 30. |
CI122C FThe area of parallelogram ABCD is 36 cm2Calculate the height of parallelogram ABEHif AB-4.2 cm.I. |
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| 31. |
Findtheonesdigitofthecubeofeachofthefollowingnumbers(b) 59(c) 96(d) 122(a) 21 |
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Answer» (a) unit's digit = 1 unit's digit of cube = 1(b) unit's digit = 9 unit's digit of cube = 1(c) unit's digit = 6 unit's digit of cube = 6(d) unit's digit = 2 unit's digit of cube = 8 chutiya kahin ka |
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| 32. |
(c) 7335+1222. Find the least numberunnce exactly.at should be subtracted from 1000 so thatbe s |
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Answer» According to Euclid's division algorithm, a=bq+r Where, a= dividend b=divisor q=quotient And r=remainder. So let a=1000, b=35, So we get, 1000= 35×28+20 Subtracting 20 from both sides, 1000–20=35×28+20–20 Thus, 980=35×28 Therefore, as seen above,980 is perfectly divisible by 35. And so,20 is the smallest number to be subtracted. |
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| 33. |
train(274) The diagonal of a rectangular ground is 60 metres more than the breadth of the ground. If the length of the groundis 30 metres more than the breadth, find the area of the ground2 hours earlier If the |
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| 34. |
lines Land M are parallel to calliOther in the figure below:PUnLQRP =800 andCRST 350what is the measure of CSPR? |
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Answer» In triangle PQR<PQR = <RST = 35°(Alternate interior angles) Now,In a triangle sum of all angles is 180° So, in triangle PQR<QPR + <PQR + <PRQ = 180°<QPR = 180 - (80 + 35)<QPR = 180 - 115 = 65° not sure but still I try....SRT=80(vertically opposite)pqr=35(alternate interior) |
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| 35. |
The suffUr tvThe perimeter of a rectangular field is 48 m. If its length is 12 m more than the breadth, findthe length and the breadth of the rectangular field. |
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Answer» my answer is not coming way |
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| 36. |
Solve If(x)=2 x_{2}-3 x-4 \text { find } f(x+2) |
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| 37. |
2sin 68° 2cot15° \_]3 tan 45°. tan 20°. tan 40°. tan 50°0s22° 5tan75° 5. Evaluate: |
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Answer» Please hit the like button if this helped you |
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| 38. |
\operatorname { tan } 20 ^ { \circ } \operatorname { tan } 40 ^ { \circ } \operatorname { tan } 60 ^ { \circ } \operatorname { tan } 80 ^ { \circ } = 3 |
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| 39. |
x_ n /5=1/8 |
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Answer» n=5/8 is the correct one n=5/8 is the right answer n= 5/8 is the right answer |
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| 40. |
\frac{2 \sin 68^{\circ}}{\cos 22}-\frac{2 \cot 15}{5 \tan 75}-\frac{3 \tan 45 \tan 20^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 70^{\circ}}{5}=1 |
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Answer» sin68= cos72tan15 = cot75tan50 = cot40tan20= cot70as these all follow compliment ruleso after putting the value2-2/5-3/52-1= 1 |
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| 41. |
i /x_—_fi -6 f p_fl + \v+ 5 |
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| 42. |
Ans. で0.25.Multiple Choice Questions: 1, 0.1 × 0.1 represents(Mental A11000(a) 102. 2.30 and 2.03 are100(a) like decimals(c) Equal(b) Unlike decimals(d) of these.3. 17 rupees 7 paisa expressed in decimal is(a) 17.70(c) 7.17Perimeter of a square of side 4.5 cm is(a) 14.5 cm(c) 20 cm176 and 176.0 are(b) 17.07(d) 17.7.(b) 18 cm(d) 45 cm |
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Answer» 8.2 change into decimals |
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| 43. |
POR is a triangle, right-angled at P. IfPQ 10 cmand PR 24 cm, find QR. |
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| 44. |
PQR is a triangle, right-angled at P. If PQ 10 cmand PR 24 cm, find QR. |
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| 45. |
1. PQR is a triangle, right-angled at P. If PQ=10 cmand PR = 24 cm, find QR. |
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Answer» so,the length of QR is =PRSO,QR=24cm the length of QR is is PR so the length of QR is =24cm QR = 26cm is right answer. please accept as best answer QR=√10^2+24^2 √100+576 =√676 =26 PQ = 10PR= 24QR = 10+ 24 100+576=676=26 so,the length of QR is =PRSO,QR=24CM pqr is a triangle right, angled at p. if PQ = 10 cm , find QR pqr is a triangle, right angled at p. and PR equal to 24 cm, find QR |
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| 46. |
1. POR is a triangle, right-angled at P. IfPQ 10 cmand PR24 cm, find QR |
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| 47. |
A = 360d the area of the shaded region in figure, where arcs drawn with centres 4, B, C anD intersect in pairs at mid-points P, Q, R and S of the sides AB, BC, CD and DA respectively of a(Use π = 3.14square ABCD of side 12 cm.APD R CSolution, ABCD is a square of side 12 em and ares d |
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| 48. |
(a) Painting a square wall of side 12 m, at the rate of rs 5per sq.cm |
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| 49. |
In an AP, ifS_{n}=n(4 n+1)), find the AP. |
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Answer» Thanks |
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| 50. |
If the side of a square is 12 cm, find its area. |
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Answer» Area of square = side² = 12² = 144 cm² |
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