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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 75351. |
In what ratio must water be mixed withmilk to gain 20% by selling themixture at cost price? |
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Answer» 100 liter milk sale has for 120 liter to make 20% profit100milk + 20 water=120 literso water:milk =1:5 |
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| 75352. |
By selling 100 toys a dealer gains the welling price of 20 toys. Find his gain percent. |
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Answer» Let the cost price of a toy be x Cost Price of 100 toys = 100x Selling 100 toys, the dear gain 20 toys: Gain = 20x Find the gain percentage: Gain Percentage = (20x ÷100x) x 100 = 20% |
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| 75353. |
A dealer marks his goods 20% more than the cost price, and offers a discount of 10%.Find his gain or loss %. |
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| 75354. |
A dealer sells a transistor at a gain of 16%. If he had sold it forRS 20 more, he would have gained20%. Find the CP. of transistor. |
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| 75355. |
LUSTRATION 17. A dealer marks his goods 20% above cost but allowsntof 10% to the custoners. Find his actual gain per cent. |
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Answer» (100+profit)/(100+add on%)*100-100=10(100+profit)/(100+20)*100-100=10100+profit/120*100=110100+profit=110*120/100profit=132-100=32% |
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| 75356. |
Ex. 110, A tradesman sells his goods at 20% discountand still makes a profit of 28%. How much % above the costprice must he mark his goods? |
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| 75357. |
After allowing a discount of 10% on the marked price, a trader still makes a gain of 17%what per cent is the marked price above the cost price? |
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Answer» thanks |
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| 75358. |
प्रश्न 52. बहुपद x-6x+11x-6 का गुणनखण्ड होगा :-(अ) (x+1) (x+2) (x+3) (ब) (x+1) (x-2) (x-3)(स) (x-1) (x-2) (x-3) (द) (x-1) (x+2) (x-3)Q. 5आप4 |
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Answer» answer to this is (x-1)(X+2)(x-3) |
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| 75359. |
A dealer offers a discount of 10% and stillmakes a profit of 25%. What price should hemark on a cooler that costs him Rs 2,000? |
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| 75360. |
A dealer buys an article for 380. At what price must he mark it so thatafter allowing a discount of 5%, he still makes a profit of 25%? |
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| 75361. |
A dealer buys a sofa set for 2,4,000. At what price must he mark it so that afterallowing a discount of 20%, he still makes a profit of 30%? |
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| 75362. |
waracit 4.1(i) (x+1): = 20x-3)(iii) (x-2)(x +1)=(x-1)(x +3).(v). (2-1)(x-3)=(r+5)(x-I)(ii) r-2x = (-2) (3-x)(iv) (x-3)(2x +1 ) =x(x +5)(vi) r+3x+1=(x-2)2(vii)(x + 2)'=2x(x-1) |
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| 75363. |
Q 27. A dealer buys an article for Rs 380. At what price must he mark it so that afterallowing a discount of 5%, he still makes a profit of 25%? |
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| 75364. |
6. A dealer offers 20% discount on the marked price of all musical instruments and still makes a profit of25%. If he gains 150 on the sale of a harmonium, then find the marked price of the instrument. |
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| 75365. |
allowing a discount of 5% on marked price a businessman still makes awhat percent is the marked price above the cost price?n marked price a businessman still makes a gain of 8%. By |
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| 75366. |
0AparallelogramABCD is given in the figure (a), in which,() In the given figure (b), ABCD is a parallelogram. If ACAB 329 and2. (<D = 115°. Find the measures of LA and <B.<ABC = 110°, calculate <x, <y and <2.115°110Quadrilaterals63 |
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| 75367. |
ๆ ) i dre uTE sin 35° = 0.5736 , sin 60° = 0.8660sin 85° = 0.99621 |
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Answer» Apologies we are currently taking questions in English language only |
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| 75368. |
sin 35°कि. Evaluate: |
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Answer» sin35/cos55=sin35/sin(90-55)=sin35/sin35=1 |
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| 75369. |
2.35ofPOLYNOMIALSdlIf the squared difference of the zeros of the quadratic polynomial f(x)-x2+ px + 45 isequal to 144, find the value of p13Ä°ratic polyn omal f(x) = x2-px + q, prove thatf the guad |
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| 75370. |
itpolynomialIf the produtlfo,the)2-6xorf(x= ar+ 6 is 4. Find the value of a16.are the zeroes of the polynomial such that α + β-6 and αβ = 4 write the polynomialeroes of the quadratic polvnomial f62nd ovic |
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Answer» Quadratic=x²-Sx+PGiven,S=6P=4 x²-6x+4 |
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| 75371. |
|ज्ञात को,11नाले भिन में अहले।3 43. जोड़ें।- लंबाई ज्ञात की।3+15+24. घटाएँ।8-45. सरल करें।6. गुणा करें।7. भाग दें।8. हल करें।मुकेश के पास 70 रु० है। भवेश के |
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Answer» If x and y are two integer x is predecessor than y.find thwe value of x_y+5 3) 3 + 1 (2/3) + 2 (1/6) = 3 + 5/3 + 13/6 = (18+10+13)/6 = 41/6 4)8 - 4 (2/7) = 8 - 30/7 = 26/7 5)6 (1/2)+ 2 (1/3)- 3 (1/4)+ 1 (2/3)= 13/2 +7/3 -13/4 +5/3= 13/4+12/3=13/4 + 4= 29/4 6) 3 (5/8) × 1 (3/5) = 29/8 × 8/5 = 29/5 7)3 (4/7) ÷ 2 (1/7) = 25/7 ÷ 15/7= 25/15= 5/3 |
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| 75372. |
When the polynomial f(x) is divided by (x-2), the remainder is (3x - 3) What is the remainderwhen (x- 1)f(x) is divided by (x-1)(x-2)61 |
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| 75373. |
polynomial f(x) 6x2-x- 2. By the method of splitting the |
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| 75374. |
7.The marked price of a computer is * 22,000. After allowing a 10% discount, adealer still makes a profit of 20%. Find the cost price of a computer. |
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| 75375. |
7. The marked price of a computer is R 28,600.After allowing a discount of 25%, the dealerstill makes a profit of 10%. Find the costprice of the computer. |
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| 75376. |
(2) Rehana purchased a scooter in the year 2015 for 260000. If its value follby 20% every year what will be the price of scooter after 2 years ?its value flsSolution: P 60000AAmount obtained after two years |
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Answer» Initial amount = 60,000 falling rate per year = 20% decrement of amount in 1st year =20 % of 60,000= 20 × 60,000/100 = 20 × 600 = 12,000 now, rest amount = 60,000 - 12,000 = 48,000 decrement of amount in 2nd year = 20% of 48,000 = 20 × 48,000/100 = 20 × 480 = 9,600 now, rest amount = 48,000 - 9,600 = 38,400 Rs hence, price of scooter after two years = 38,400 Rs. Please like the solution 👍 ✔️👍 |
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| 75377. |
. If the highest common factor of numbers 408and 1032 is expressed in the form of 1032x408 x 5, then find the value of x. |
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Answer» If the HCF of 408 and 1032 can be written as By Euclid 's division algorithm, 1032 = 408×2 + 216 408 = 216×1 + 192 216 = 192×1 + 24 192 = 24×8 + 0 Since the remainder becomes 0 here, so HCF of 408 and 1032 is 24 Now, 1032x - 408*5 = HCF of these numbers => 1032x - 2040 = 24 => 1032x = 24+2040 => x = 2064/1032 => x = 2 So value of x is 2. Like my answer if you find it useful! |
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| 75378. |
ail market, fruit vendors were selling mangoes kept ina retakes contained varying number ofangoes according to the number of boxes.angoes. The following was the ishation ofNumber of mangoes 50-52 53 ss56.58| 59.61|62.64Number of boxes1s1101351155 |
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| 75379. |
( \frac 5 \sin 35 ^ \circ \cos 55 ^ \circ ) %2B ( \frac \cos 55 ^ \circ 2 \sin 35 ^ \circ ) - 2 \cos 60 ^ \circ |
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Answer» 5sin(90-55)/cos55+cos55/2sin(90-55)-2*1/2=5cos55/cos55+cos55/2cos55-1=5+1/2-14+1/2=9/2 |
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| 75380. |
8 Obtain all the zeros o the polynomial f(x) 3x1 + 6x3 - 22 - 10x -5, iftwoofits zeros areaand3 |
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| 75381. |
isavigs12. The number of cards collected by Rina, Mina, and Tina are in the ratio 1:3: 5 respectively. If Rina hasa collection of 45 cards, find the number of cards with Mina and Tina. |
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Answer» rina = 5, mina = 15, Tina = 25 |
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| 75382. |
The cost of a computer is 27540 How much will 18 such computer cost? |
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Answer» x square - minus 81 ka factor Answer : 27540× 18and thefinal answer is : 495720 4,95,720 is the correct answer cost of 1 comp=27540cost of 18 comp=18×27540 =495720 computer = 27, 540; 18 x 27540 = 495, 720 Answer 27540x18=495,720 27540 X 18 = 495720I hope ot helps 495720 is the correct answer multiply by 18 with 27540 495720 this us direct answer 18 computers cost 495720 rupees |
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| 75383. |
(1-x)?29 Find the mean , variatiec and standard deviation for the followmg.data:-70-80 80-90 90-100Marks 30-40740 50 50-60 60-708 3 T212 157No. of 3students+-4KD- 42 K3)2KK |
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| 75384. |
\left. \begin{array} { l l } { \text { Simplify. } } & { ( 1 ) ( x - 2 y + 3 ) ^ { 2 } + ( x + 2 y - 3 ) ^ { 2 } } \\ { ( 2 ) ( 3 k - 4 r - 2 m ) ^ { 2 } - ( 3 k + 4 r - 2 m ) ^ { 2 } } & { ( 3 ) ( 7 a - 6 b + 5 c ) ^ { 2 } + ( 7 a + 6 b - 5 } \end{array} \right. |
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| 75385. |
The efficiency of a machine decreases by 2 units every year. If theefficiency of a new machine is 45 units, calculate its efficiencyafter 8 years. |
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Answer» efficiency45-16=29 units efficiency45-16=29 units efficiency45-16=29units |
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| 75386. |
-cot(45^circ)^2 %2B sec(45^circ)^2 - sin(30^circ)^2 - sin(60^circ)^2 |
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Answer» sec^2 45 - cot^2 45 - sin^2 30 - sin^2 60 = sec^2 45 - tan^2 (90 - 45) - sin^2 30 - cos^2(90 - 60) = (sec^2 45 - tan^2 45) - (sin^2 30 + cos^2 30) = 1 - 1 = 0 O is the right answer |
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| 75387. |
Dinesh went from place A to place B and fromthere to place C. A is 7.5 km from B and B is A12.7 km from C.Ayub went from place A to placeD and from there to place C. D is 9.3 km from Aand C is 11.8 km from D. Who travelled moreand by how much?7.ava 50 g mangoes. Sarala |
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| 75388. |
Kanchanis thinking of a 5-digit number.The number is a multiple of 5. The ten'splace digit is the highest common factor of16 and 24. The hundred's place digit is thesmallest odd number. The ten thousand'splace digit is an even prime number. Thethousand's place digit is the third multipleof the sum of the ten thousand's placeand hundred's place digit. There are only2 even digits in the number. What numberis Kanchan thinking of? |
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| 75389. |
+1, tan βtan β-_ t -, then α + β is equal2m41, tIf tan α-,m +1toINCERT Exemplar)4 |
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| 75390. |
(sin(27^circ)^2 %2B sin(63^circ)^2)/(cos(17^circ)^2 %2B cos(73^circ)^2) |
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Answer» (i) sin^2 (63)+ sin^2(27)/ cos^2(17)+ sin^2(73)= cos^2(90-63)+ Sin^2(27)/ cos^2( 17)+ sin^2(90-73)= cos^2(17)+ sin^2(27)/ cos^2(17)+sin^2(27/=1. (ii) 1. 12.9921488722. 1 (ii)sin25 cos65+ cos25 sin 75=sin25 cos(90-65)+ cos(90'-17)+ cos^2(73)= sin25cos25+ sin^2(73) cos^2(73)=1 1)1 2 )1 is the correct answer of the given question. |
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| 75391. |
13. Given that 2 is a zero of the cumbic polynomial 6x3+ /2r2-10x -4/2, find itsINCERT EXEMPLAR]other two zeroes. |
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Answer» √2 is zero of 6x³ +√2x² -10x -4√2 , So, (x - √2) is a factor of 6x³ + √2x² -10x -4√2 . 6x³ + √2x² -10x -4√2 = 6x³ -6√2x² + 7√2x² -14x + 4x - 4√2= 6x²(x - √2) + 7√2x(x -2) + 4(x -√2) = {6x² + 7√2x + 4}(x - √2)= {6x² + 3√2x + 4√2x + 4}(x -√2)= {3√2x(√2x + 1) + 4(√2x +1)}(x -√2)= (3√2x +4)(√2x +1)(x - √2) Hence, two other zeros are -4/3√2, and -1/√2 |
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| 75392. |
A class test in mathematics was conducted for class VI of a school. Following dgives marks (out of 60) of students:istribution20-301250-603Marks30-40100-1010-2040-50Number ofstudents[Ans. Mean - 23.5Find the mean of the marks obtained. |
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| 75393. |
rebelow how one part of the writingPODLOW show thetoacks intandingpHOUGH WORKITina starts walking in the clockwise direction. Shetakes the third track on her left. Then, she takes thefirst right turnTowards which direction is Tina walking now?A. South WestB. North-EastC. EastD. North |
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Answer» I think south west is the correct answer south west is the correct answer south west is the correct answer |
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| 75394. |
30r 2l 09 x " |
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Answer» x- 3/x=1/2 x²-3=x/2 2x²-6=x 2x²-x-6=0 2x²-4x+3x-6=0 2x(x-2)+3(x-2)=0 (x-2)(2x+3)=0 x=2,-3/2 |
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| 75395. |
63. यदि 5A = 6B तथा 6A = 4C, B: Cका अनुपात है |
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Answer» 5A = 6B A = 6B/56A = 4C6(6B/5) = 4C 36B/5 = 4C B/C = 4 × (5/36) = 5/9B:C = 5:9 B:C=5:9 is the right answer. |
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| 75396. |
1. Find the square of (V5a+6b) |
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| 75397. |
2tan AIf cosAs-xī, then find the values of sinA and tanA in terms of x. |
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Answer» we know,cos A = B/H = 2x/1 + x^2 Using pythagoras theoram H^2 = P^2 + B^2 (1 + x^2) = P^2 + 4x^2 1 + x^4 + 2x^2 = P^2 + 4x^2 P^2 = 1 + x^4 - 2x^2 P^2 = (1 - x^2)^2 P = (1 - x^2) Therefore,sin A = P/H = (1 - x^2)/(1 + x^2) tan A = P/B = (1 - x^2)/2x |
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| 75398. |
Describe the multiplication of 5a-6b and 3a-2b? |
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Answer» (5a-6b)(3a-2b)5a(3a-2b)-6b(3a-2b)=15a²-10ab-18ab+12b² =15a²-28ab+12b² |
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| 75399. |
3(-ly: (-1)" sing [sin(-8)s-sind: 6-1)sind-(-D)nsin 4 sin42EXERCISE 5.11. frrfefea erer(Prove the following identities.)(i) cose.tane sine(ii) tan-sin0VI-sin20(iv) sin2A+cOs2A.sin B sin B +cos-B.sin2 A2. frs f (Prove that)(ii) cosec-0-cot 0 coscc20+co0(iii) (I + tan0)2-sec282 2tan®(iv) (sin0 + cos0) (1-sin0.cos0) sin30+ cos30(v) sin 0+cost0 1-2sin20.cos0(vi) (cosec θ-sie) (sec0-cos0) (tane + cot0) : 1(viii) ( 1-sin®-cos0)2-( 1-sin® (1-cos)(ix) (I +sin+cos0)2 2(1sine) (1 + cos8)(x) cosag.cos2β-sinzg.sin2pecosa-sin2ß3. frai fs (Prove that):(İ)tanB+ cot0-sec0.cosec0COSACOSA2secA(ii)1-sinA + 1 + sinAcos?a-cosB8cos2α.cos2ß tan?β-tana(iv)1-sin6+sine 2sec0.tan |
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| 75400. |
(A) tan 90°(B) I(ii)i sin 2A 2 sin A is true when(A) 0°(B) 30°2 tan 30°(iv) - tan 30(A) cos 60(B) sin 6 |
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Answer» sin 2A = 2 sinA 2cosA sinA = 2 sin A cos A = 1 So, A = 0 |
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