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. |
In how many ways can 6 boys and 4 girls sit in a noso that row boy is between two girls? |
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Answer» But mam answer is 120960 |
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| 2. |
(ii) 3x2-2, 6x+2-0 |
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| 3. |
A society collected rupees 2916. each member contributed as many rupees as there were members. how many members were there? |
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Answer» number of members = amount they paid number of members = (2916)^1/2 = 54 (ANS) let no of member = xcontribute = xx.x= 2916x^2= 2916x= √2916= 54 so no of member = 54 number of members = amount they paidnumber of. members =(2916) ^1/3 = 54(ans) number of members =amount they paid number of members =(2916)^1/2=54(ANS) |
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| 4. |
(1/2)^1990 - 1/224244276420752836764225234605016012504358926189198281343289597036282472928197326925128805921192630843446477468552238822938890422043565494419127218001299450271738260049438046876345271773906337639509623027467746571839092702434352273334603566485285847599423821402660935173719979698483030565500320746035806018250329900360270518754647431265457146290888235352400395931067628575238297133273986897971226334603919138662883629154339413785945394843463248538165799675590023282333901921073578201484535072688731606944102393718214874078806223475453231462725148015395752295772517189107114786753838776879267287400448=2^x |
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| 5. |
Find HCF and LCM of 404 and 96 and verify that HCF xLCMgiven numbers.Product of the two |
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Answer» The H.C.F. of 96and 404is |
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| 6. |
Ram has 60 more one rupee coins than five-rupee coins. The total value of the money is ? 360.How many five rupee coins Ram has?7.8. Savita buys postage stamps of denomination of 2 1,25 and 10 in the ratio 2:5:3. She pays 741 tothe cashier Find the number of postage stamps of each kind |
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| 7. |
A man purchased a watch for Rs.400 and sold it at a gain of 20% on the selling price. The Selling,price of the watch is: |
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Answer» CP = 400 GAIN = 20% SP = 400*1.2 = 480 |
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| 8. |
Find the toloatural unbes |
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Answer» The first ten natural number are 1,2,3,4,5,6,7,8,9,10. thank you |
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| 9. |
(vi)given a = 4, d=2, S =-14, find n and a.tolo |
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Answer» Sn = n/2(2a+(n+1)d) Now an= a+(n-1)d4= a+(n-1)a+n = 5 An=a+(n-1)d=4a+(n-1)2=4 (substitute d=2)a+2n-2=4a+2n=6a=6-2n Sn=n/2(2a+(n-1)d)-14=n/2(2(6-2n)+(n-1)2) (substitute a=6-2n,d=2)-14=n/2(12-4n+2n-2)-14=n/2(10-2n)-14=n(5-n)5n-n^2=-14n^2-5n-14=0n^2-7n+2n-14=06-2n(n-7)+2(n-7)=0(n-7)(n+2)=0n=7,-2 n can't be negativetherefore n=7 a=6-2n (substitute n=7)a=6-2(7)=6-14=-8 |
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| 10. |
EXERCISE 2.4Determine which of the following polynomials has (x+ 1) a fact(0) x3+x2 +x+ 1(i) 4+x+x2+x+1 |
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Answer» thanks bro |
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| 11. |
If the mean of x,, X2 is 7.5 and the meanof x,, X2, X is 8, then the value of x is |
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Answer» (8*3)-(7.5*2)=9. Done |
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| 12. |
Q1. Find HCF and LCM of 72, 126, 156. Also verify that HCF XLCMProduct of three numbers.ware |
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Answer» L cm 72, 126,156,; 72=2×2×2×3×3; 156=2×3×3×7; 156=2×2×3×13; L cm =2×2×2×3×3×7×13=8×9×91=6056; Hcf=2×3=6; L cm x Hcf=6057 x 6=36+0+30×42=45036 |
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| 13. |
Use suitable identities to find the followingproducts,i, at 4) (atla) |
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Answer» (x+4) (x+10) x^2+10x+4x+40=x^2+14x+40 6x is the right answer (x+4) (x+10)xdhdjzbsjxhakbxjejzbejwhsjsjjdbd |
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| 14. |
ob d fern eteT6 on. Rndĺśhe height of each |
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Answer» please like my answer if you find it useful |
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| 15. |
16. I have a total ofき300 in coins of denomination1,乏2 and5. The number of2coins is 3 times the number of 5 coins. The total number of coins is 160. How manycoins of each denomination are with me? |
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Answer» Thankyou so much |
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| 16. |
15. I have a total of Rs 300 in coins of denomination Re 1, Rs 2 and Rs 5. Thoenumber of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160. Howmany coins of each denomination are with me? |
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Answer» Given, total value of Rs = Rs 300Total number of coins = 160Coins of denomination = Re 1, Rs 2 and Rs 5Number of Rs 2 coins = 3 x number Rs 5 coinsLet the number of coins of Rs 5 = mSince, the number coins of Rs 2 is 3 times of the number of coins of Rs 5Therefore, number of coins of Rs 2=m×3=3mNow, Number of coins of Re 1 = Total number of coins – (Number of Rs 5 coins + Number of Rs 2 coins)Therefore,Number of coins of Re 1= 160 – (m+3m) = 160 – 4mTotal Rs = (Re 1 x Number of Re 1 coins) + (Rs 2 x Number of Rs 2 coins) + (Rs 5 x Number of Rs 5 coins)⇒ 300 = [1×(160–4m)] + (2×3m) + (5×m)⇒ 300 = (160–4m) + 6m + 5m⇒ 300 = 160 – 4m + 6m + 5m⇒ 300 = 160 – 4m + 11m⇒ 300 = 160 + 7mAfter transposing 160 to LHS, we get300 – 160 = 7m⇒ 140 = 7mAfter dividing both sides by 7, we get 1407 = 7m7Or, m = 20Thus, number of coins of Rs 5 = 20Now, since, number of coins of Re 1=160–4mThus, by substituting the value of m, we getNumber of coins of Re 1= 160 – (4×20) = 160 – 80 = 80Now, number coins of Rs 2 = 3mThus, by substituting the value of m, we getNumber of coins of Rs 2= 3m = 3×20 = 60Therefore,Number of coins of Re 1 = 80Number of coins of Rs 2 = 60Number of coins of Rs 5 = 20 any short method |
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| 17. |
के एक लाख ३को को परस्पर बदल56.48. यदि संख्या 132943 के एक लाहजारवे स्थान वाले अंको को परसदिया जाए, तो नई संख्या मूल संख्या मेंकितनी अधिक हो जाएगी?({} 180000(2) 310000(3) 2300{4) 33000fern Tha |
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Answer» 310000 is a correct answer |
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| 18. |
15. I have a total of Rs 300 in coins of denomination Re 1, Rs 2 and Rs 5. TheC+number of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160. Howmany coins of each denomination are with me? |
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Answer» Given, total value of Rs = Rs 300Total number of coins = 160Coins of denomination = Re 1, Rs 2 and Rs 5Number of Rs 2 coins = 3 x number Rs 5 coins Let the number of coins of Rs 5 = mSince, the number coins of Rs 2 is 3 times of the number of coins of Rs 5Therefore, number of coins of Rs 2=m×3=3mNow, Number of coins of Re 1 = Total number of coins – (Number of Rs 5 coins + Number of Rs 2 coins)Therefore,Number of coins of Re 1=160–(m+3m)=160–4mTotal Rs = (Re 1 x Number of Re 1 coins) + (Rs 2 x Number of Rs 2 coins) + (Rs 5 x Number of Rs 5 coins)⇒300=[1×(160–4m)]+(2×3m)+(5×m)⇒300=(160–4m)+6m+5m⇒300=160–4m+6m+5m⇒300=160–4m+11m⇒300=160+7m After transposing 160 to LHS, we get300–160=7m⇒140=7m After dividing both sides by 7, we get 140/7=7m/7 Or,m=20Thus, number of coins of Rs 5 = 20Now, since, number of coins of Re 1=160–4mThus, by substituting the value of m, we getNumber of coins of Re 1=160–(4×20)=160–80=80Now, number coins of Rs 2 = 3m Thus, by substituting the value of m, we getNumber of coins of Rs 2=3m=3×20=60Therefore,Number of coins of Re 1 = 80Number of coins of Rs 2 = 60Number of coins of Rs 5 = 20 any other short method plzz.. |
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| 19. |
SHAMSNYg s ¢ — g 99500 922 |
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Answer» Like my answer if you find it useful! |
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| 20. |
2८* - 9 - 922 +62 का गुणनखण्ड है |
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Answer» x² - y² - 9z² + 6yz x² - ( y² + 9z² - 6yz) x² - ( y - 3z)² ( x - ( y - 3z)) ( x + ( y - 3z)) ( x - y + 3z) ( x + y - 3z) |
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| 21. |
Eramêne the following function ble) dovContinuite, at 2:10)2 52-4 LOKALI0 922-20 KAL2 |
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| 22. |
25Differentiate ittent (922)Llan |
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| 23. |
What is to be added to 63.58 to get 922 |
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Answer» 92 - x =63.58x=92-63.58x=28.42 28.42 is to be added to get 92 28.42 the correct answer of the given question 92-63.58=28.42so 28.42 will be added in 63.48 to get 92. 63. 58 + x = 92; X = 92 - 63. 58 = 28. 42 28.42 is correct answer 28.42 us correct anseer 28.42 is the correct answer |
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| 24. |
So, we obtain the lolltsample 25: Factorise : K2ISolution : Here, we have8 27- 18z- (2x)y (B)-3(200x3:)(22ty(2x + y + 32) (4x2 + y,2 + 922-EXERCISE 2.5to find the following productsL. Use suitable identities(0 (a+4)(r+ 10)(nMx +8) (x-10) |
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| 25. |
+ के लिए निम्नलिखित समीकरण को हल कीजिए:फल - 909 न पु)ज न (2017 59५7 20) न 0 |
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Answer» hit like if you find it useful |
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| 26. |
Factorise: (x+x +4(x2+x)-12 |
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| 27. |
q(x) =2+x-x2;x=2 |
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Answer» q(2) = 2 + 2 - 2^2 = 4 - 4 = 0 |
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| 28. |
x+4=66 |
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Answer» x+4=66x=66-4x=62 is the answer |
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| 29. |
he ratio of men to women passengers in an aeroplane is 4 : 7. There are14. 17 women, how many are men? How many total number of passengers are there? |
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Answer» men and women ratio=4:7total ratio=4+7=11such 7women in 11passengerso 1women 11/7so 147 women. 11*147/7=231ans |
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| 30. |
Fund the hcf and lcm of 72 126 and 168 |
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| 31. |
(गण) 254 हा22=2" |
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Answer» hit like if you find it useful |
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| 32. |
Ifx= 3-252, findthe value of dette |
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| 33. |
3 x 4² x 29Igbox 26 |
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Answer» 27 is the correct answer of this question the right Ans is 9/16. |
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| 34. |
[ w7 3 ) -13 -र7ं 279) | —+ = |t —— = g | नए की ना15 : 5: 22 K =3 8 |
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| 35. |
factorine the following: 922 6x +1-254 |
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Answer» 3*3,6x+1-5*5y9-1+6x10y8+6x10y 9x^2, 6 x+1-25y; 9x^2-6x+1-24y |
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| 36. |
x3-4x2-x+1 = (x+29 |
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| 37. |
X2-(x-29=32 |
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Answer» If you find this solution helpful, Please like it. |
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| 38. |
66 x 29 (e) |
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Answer» 67 * 99 = 6633 is the answer thanks for the answer.I was just checking this app how it works.thanks 67*99 =6633 is the product |
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| 39. |
x का मान ज्ञात कीजिए,X+29 = 411 |
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Answer» x +29 =411411-29=xx=411-29=382 and. x+29=411x=411-29x=382 382 is right answer hi X + 29 = 411 x=411-29x=382 x+29=411x = 411-29x = 382 is write answer x+29=411x=411-29x=382 x+29=411x=411-29x=411 x+29=411x=411-29x=382 this is correct answer x +29= 411; x=411-29= 382 x + 29 = 411x = 411 - 29x = 382Thus,the value of x in the given equation is 382. in given equation x is equal to 382 x + 29 = 411 x = 411 - 29 x = 382. X = 382. 382 this is a answer of these question 382 is the answer of your given question 382 is right answers x+29=411 x=411-29 x=382 in right answer x=411-29 X=382 in the right answer the right answer is 382 x + 29 = 411x = 411 - 29x = 382Therefore, the value of x in the given equation is 382. x+29=411x=411-29x=382.is right answer please like x = 382 is right answer |
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| 40. |
'ठै1-.. यदि ४. >, का माध्य 7.5 तथा X,» X,, X, S W7 § “लितो +, का मान है |
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Answer» thanks |
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| 41. |
4. Solve for x:29 |
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Answer» Like my answer if you find it useful |
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| 42. |
10 Find the digit at the ones place of the product 252 x 168 x 72 without working out theproblem.Com |
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Answer» 6×5×90 at ones digit 0 at its ones place is the right answer 6x5x90 at one digit |
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| 43. |
254 x ^ 2 - 4 a ^ 2 x %2B ( a ^ 4 - b ^ 4 ) = 0 |
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| 44. |
In the figure, seg AB and seg CD intersect at thepoint O. OA=15 cm, OB=12 cm, OC= 10 cm,OD= 8 cm. State whether ∆AOC and ∆BOD aresimilar or not. If so, by which test? |
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| 45. |
Simplify and write in the exponential form: 254 x 55 |
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Answer» 25⁴ × 5⁵ = (5²)⁴ × 5⁵ = 5⁸ × 5⁵ = 5¹³ (∵ a^m a^n = a^(m + n)) |
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| 46. |
divisde the following 1. 48x×xy×y×y×y×y by 250 y×y×y×y |
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Answer» Given question is -48*x^2*y^5/250y^4=24x^2*y/125 Hence answer is 24x^2y/125 |
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| 47. |
-3v-3x=155v+X=-29 |
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Answer» -3v-3x = 15 => -3(v+x) = 15 => (v+x) = 15/-3 = -5 => v+x = -5 now, 5v+x = -29 => 4v+(v+x) = -29.... ( now, put x +v = -5)=> 4v+(-5) = -29=> 4v = -29+5 = -24=> v = -24/4 = -6 now, x = -5-v = -5-(-6) = 1 |
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| 48. |
ગાકાર શોધો.(f) * (242) x (4a26) |
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Answer» 2×4×a^2×a^22×a^26=8a^2+22+26=8a^50 8a⁵⁰ is the solution 8a^50 is the correct answer |
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| 49. |
242 -15oc-h |
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Answer» x+2=15x=15-2=13x=13 |
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| 50. |
In the adjoining figure, it is given that ABⅡ CD,"ABO=50° and<CDO = 40°.Find the measure of ZBOD50°Hint. Through O draw EOF II AB40° |
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