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| 6101. |
13. li, sinc)cosO, then find the value of 2 tan θ + cos θ→2LE cos Afind the value of, 4+4 tan-4 |
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Answer» It is given that , sinx = cosx we know, at x = π/4 , sinx = cosx = 1/√2 so, x = π/4 or 45° now, 2tanx + cos²x = 2tanπ/4 + cos²π/4 = 2 × 1 + (1/√2)² = 2 + 1/2 = 5/2 |
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| 6102. |
5150VE10. (a) Ifx (4-5-3find the value of x-1b) Ifx 2Ix (-4), find the value of-6(32Fiul ihe value of |
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| 6103. |
(iii) |-|446+8(a) Ifx = (4 1-5-1)(3)find the value of,-.(b) If x = ili10.15| |x (-4), find the value of-6. |
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| 6104. |
- v oJb ¢b* c’|=abe(a~bXb-c)c-a)b ¢ |
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| 6105. |
oj thze ust0.5.Solve the equations-4+(-1)+2+¡+x= 437. |
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| 6106. |
Oj |
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Answer» x+y = 7 and 8x + 7y =9 so x =7-y , putting the value of x in 2nd equation8[7-y] + 7y =9 , 56 -8y +7y =0 , 56 = y and x = 7 - 56 =-49 |
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| 6107. |
ee angles ataiangle an e in theo:3 Fird the m easune oj eacA angle |
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Answer» let the ratio be x so angle are x , 2x and 3x so sum of angles in triangle 180x+2x+3x = 1806x = 180x= 180/6 = 30so angles are 30 , 60 and 90 |
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| 6108. |
12) A candidate who gets 20% marks fails by 10 marks but another candidatewho gets 42 marks gets 12% more than the passing marks.Find the maximum marks |
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Answer» Thank u |
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| 6109. |
To receive Grde ·A, in a course, one must obtain an average of 90 marks omore in five examinations (each of 100 marks). If Sunita's1examinations are 87,92, 94 and 95, find minimum marks that Sunita must obtainin fifth examination to get grade A' in the course.23. nd l |
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| 6110. |
A candidate who gets 36%,another candidate, who gets 43% narks, gets 18 more marks than thenarks in the examination fails by 24 marks butinimum pass marks. Find the maximum marks and the percentage of passmarks. |
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Answer» First candidate = 36% × x + 24Second candidate = 43% × x - 18First candidate = Second candidate36/100 × x + 24 = 43/100 × x - 189x/25 + 24 = 43x/100 - 189x/25 - 43x/100 = -24 - 1836x/100 - 43x/100 = -4236x - 43x/100 = -42-7x/100 = -42x = -42 × -100/7x = 600Maximum marks = 60036% × x + 24 = y36/100 × 600 + 24 = y216 + 24 = y240 = yPassing marks = 240 |
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| 6111. |
104Therefore each man gets 2 kgEXERCISE 3.41. Divide(a) 15 - 20 (b) 48 – 2624(C) 2-5160Oivala ow€2013(6) 1-2-2. Sunita bought 27 meters of rope. She cut the rop |
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| 6112. |
If x -3 is one root of the quadratic equation.x?-2kr-6-0, then find the value of k. |
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| 6113. |
Ifx= 3 is one root of the quadratic equation x2-2kr-6-0, then tind the value of k. |
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Answer» As x = 3 is root of equation x^2 - 2kx - 6 = 0 Then,For x = 3 equation will be equal to 0 Hence,(3)^2 - 2k*3 - 6 = 09 - 6k - 6 = 06k = 3k = 3/6 = 1/2 |
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| 6114. |
Ifx-3 is one root of the quadratic equation x-2kr-6 = 0, then find the value of k.the |
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Answer» If x=3 is one root of the quadratic equation,Let p(x)=x²-2kx-6=0 =>p(3)=3²-(2×k×3)-6=0=>9-6-6k=0=>6k=3=>k=1/2 |
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| 6115. |
an examination, a student must get 36% marks to pass. Rachit gets 180 marks and fails by 36 marks.Find the maximum marks. |
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Answer» Rachit gets 180 marks and fails by 36 marks. So passing marks is 180+36=216 Let the total marks be x36% of x = 216 Therefore, x=(216 X 100) divided by 36=600 Therefore, the maximum marks is 600 Like if you find it useful |
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| 6116. |
Hari bought 20 kg of rice at32 per kg. He mixed the twoer cent in the whole36 per kg and 25 kg of rice atvarieties and sold the mixture at 38 per kg. Find his gain ptransaction |
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Answer» Cost of 1kg rice = ₹36Cost of 20kg rice = ₹36 × 20 = ₹720 Cost of 1kg rice = ₹32Cost of 25kg rice = ₹32 × 25 = ₹800 Total weight = 45kg Total cost = ₹1520Selling price of 45kg rice = ₹38 × 45 = ₹1710 Gain = 1710 - 1520 = 190 Gain% = (190/1520) × 100 = 12.5% |
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| 6117. |
6. Hari bought 20 kg of rice at * 36 per kg and 25 kg of rice at 32 per kg. He mixed the twovarieties and sold the mixture at * 38 per kg. Find his gain per cent in the wholetransaction. |
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Answer» Solution :- Cost of 1 kg rice = Rs. 36 per kgCost of 20 kg rice = 36*20 = Rs. 720 Cost of 1 kg rice = Rs. 32 per kgCost of 25 kg rice = 32*25 = Rs. 800 Total quantity of rice = 20 + 25 = 45 kg Cost of 45 kg rice = 720 + 800 = Rs. 1520 Selling price of the new mixture = Rs. 38 per kg selling price of 45 kg rice = 38*45 = Rs. 1710 Gain = 1710 - 1520 = Rs. 190 Gain = Rs. 190 Gain % = (Gain*100)/cost price ⇒ (190*100)/1520 ⇒ 19000/1520 = 12.5 % So, on whole transaction, Hari got profit of 12.5 % |
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| 6118. |
6. Hari bought 20 kg of rice at * 36 per kg and 25 kg of rice at 32 per kg. He mixed the twovarieties and sold the mixture at 38 per kg. Find his gain per cent in the wholetransaction. |
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Answer» cp= 20×36 +25×32= 720+800=1520then sp= (20+25)×38 =45×38=1710 profit= 190profit%= 190×100/1520 =19000/1520 =12.5% |
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| 6119. |
6.D- TIUSOU + 2600) = ? 12900.Hari bought 20 kg of rice at 18 per kg and 25 kg of rice at 16 per kg. He mixed the twovarieties and sold the mixture at 19 per kg. Find his gain per cent in the whole transaction.Coffenti 50 werd was mit |
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| 6120. |
5 Prem solda transistor to Sudhir at a gain of 10% and Sudhir sold it to Hari at a gain of 15%. Prem habought it for500, what did it cost to Hari? |
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Answer» C.P to Prem = Rs. 500Gain = 10%S.P = C.P*(100+G)/100 = 500*110/100 = Rs.550 C.P to Sudhir = Rs. 550Gain = 15%S.P = C.P*(100+G)/100 = 550*115/100 = 632.5 C.P to Hari = Rs. 632.5 |
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| 6121. |
যায়িত্ব9. log (log8. tan'x603 |
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Answer» the correct answer is C |
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| 6122. |
out of a class of 45 students, five were absent, 30% of the remaining had failed to do homework. Find the number of students who did the homework. |
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Answer» if 5 were absenthence45-5=40students30% failed to do homework left 70% hence 40*70/100=28did homework. |
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| 6123. |
Sum of the roots of a quadratic equation is doubie their product. Fin kif the equation is x2 -4kx +k+ 3 0 |
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Answer» Let α and β are the roots of quadratic equation x² - 4kx + k + 3 = 0∴sum of roots = - coefficient of x²/coefficient of x α + β = -(-4k)/1 = 4k Product of roots = constant/coefficient of x²αβ = (k + 3)/1 = (k + 3) A/C to question, sum of roots = 2 × product of roots (α + β ) = 2αβ⇒4k = 2(k + 3) ⇒ 4k = 2k + 6 ⇒ 4k - 2k = 6 ⇒ 2k = 6 ⇒ k = 3 Hence, answer is k = 3 Like my answer if you find it useful! |
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| 6124. |
the den |
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Answer» (16 sqrt2+1)/8 rationalizing it gives (16×16×2-1)/8(16 sqrt2+1)511/8(16 sqrt2+1) |
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| 6125. |
If x -3 is one root of the quadratic equation x^{2}-2 k x-6=0, then find the value of k. |
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| 6126. |
If x= 3 is one root of the quadratic equationx^{2}-2 k x-6=0, then find the value of k. |
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Answer» Polynomial - x2 - 2kx - 6 =0P (3)=(3)2 - 2(3)k =6=9 - 6k =6=-6k =-3=k =1/2 |
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| 6127. |
Ifx = 3 is one root of the quadratic equation x^{2}-2 k x-6=0, then find the value of k. |
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Answer» Given it's one root of the equation x² - 2kx - 6 = 0 one of its root x = 3 for finding the value of k. Put the value of x in this equation. (3)² - 2k(3) - 6 = 0 9 - 6k - 6 = 0 6k = 3 k = 3/6 k = 1/2 Hence, the value of k = 1/2 |
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| 6128. |
Alka bought rice for 56.25 and sold it at a proft of R 11,.25. What is the selling priceof the rice? |
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Answer» CP = ₹ 56.25Profit = ₹11.25SP = CP + profit = 56.25+11.25 =₹ 67.5 thanks |
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| 6129. |
Find the gain or loss in the followingi) C.P =750, overhead expenses-50 : Profit=, 80400 ; S.PC.PC.P =乏46000, Overheads-7 4000 ; S.P =T60000(iii)460 |
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Answer» 1. C.P = 750Overhead charges = 50Therefore total C.P = 750+50=800Profit =₹80Therefore S.P = 800+80 = 880Therefore profit% = 880/800= 1.1%2. CP = rs 400SP = rs 460CP < SPso, there is profit Now profit = SP - CPprofit = 460 - 400profit = rs 60now,profit % = [ profit/CP ] × 100profit % = [ 60/400 ]×100profit % = 15%3. CP=46000OVERHEADS=4000TOTAL CP=46000+4000=50000SP=60000PROFIT=10000PROFIT%=10000/50000×100=20ANS=20% thanks yar 1ka 10 % |
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| 6130. |
Hari bought two fans for 2,400 each. He sold one at a loss of 10%and the other at a profit of 15%. Find the selling price of each fanand find also the total profit or loss. |
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Answer» what is question? i am unable to see your question please sent it perfectly. selling price of 1st fan is 2160 and 2nd fan is2760 and je had a profit of120 rupees Given CP = 2400 Loss = 10%. We know that SP = 100-Loss%/100 * CP = (100 -10)/100 * 2400 = 90/100 * 2400 = 2160 Given Profit = 15%. SP = 115/100 * 2400 = 2760. Total CP = 2400 + 2400 = 4800. Total SP = 2160 + 2760 = 4920. Total Profit = 4920 - 4800 = 120 rupees. |
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| 6131. |
A dealer is selling an article at a discount of 15%. Find:(i) the selling price if the marked price is ₹500(ii) the cost price if he makes 25% profit. |
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| 6132. |
5. A retailer buys a radio for Rs 225. His overhead expenses are Rs 15. If he sells theradio for Rs 300, determine his profit percent. |
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Answer» so , profit = 300-250 = 50Now ,percentage profit = 50/250×100 = 20% cp of radia=225overhead expenses=15therefore Total CP=225+15=240SP of radio=300Profit=SP-Total CP=300-240=60ThereforeProfit%=(profit/total cp)x100 =(60/240)x100 =25% CP of radio=225overhead expenses=15Total CP=225+15=240SP of radio =300Profit=SP-Total CP=300-240=60Profit %=(profit/total CP)*100(60/240)*100=25% |
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| 6133. |
17. Aman buys an article for150 and makes overhead expenses which are 10% of the costprice. At what price must he sell it to gain 20%?(d) チ208 |
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| 6134. |
Find Selling PriceCast Price = 15000Lass 30%Selling Price ? |
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Answer» selling price..11050 selling price of this question is 10500 15000*30/1000=450.15000-450=14550 Cost Price=15000/-Loss=30%Selling price=19500/- Cost Price=15000/-Loss=30%Selling price=4500/- 10500 hoga bhai iska answer Your cost price is 15000 rupees only.You loss 30 percentage.Then your selling price is 10500. sepling price is equal to 10500 spelling prise 10500 selling price equal 10500 selling price..... 10500 Selling price is 10500 cost price 15000loss30% - 4500 selling price= 10500 11500/-30% 15000 is 4500/- 15000-4500 = 11500 answer Selling price is 11050 cost=15000loss=30% yani (4500)selling price = 10500 cost price =15000 loss=30%selling price=15000×30/100 =1500×3 =4500 selling price of this question 4500 S.P = Cost price - loss percentage.S.P = 15000- 30% = 10500 sellong price 45000 hogi the selling price will be 10500 the answere of this question is 4500 iska answer 15000*30/100=4500 cost price=15000loss=30%selling price=?15000÷100%=150150×30=6500loss=650015000-6500=8500selling price=8500 iska selling price 10500 hoga 10500 selling price hai 10500 selling prise hoga selling price of this question is 10500 the right answer cost price = 15000loss= 30%selling price = 4500 15000*30/100=4500,15000-4500=11500 selling price 10500 hoga 15000×30/100=4500selling price=15000-4500=10500 15000×70÷100=10500 hoga costs prize ok 👌👌👌👌👌👌👌👌👌👌👌👌👌 10500 this is a right answer 4500is right answer hai selling price will be 45,00 answer for this question is 9500 Cp=15000loos=30%SP=15000×70/100SP =10500answer Cp=15000loos=30%SP=15000×70/100SP =10500 15000×30÷100=4500 answer cost price=15000loss=30%selling price= 15000-(15000*30/100)= 10,500 selling price=10500 answer selling price (4500) selling price= 15000-30%of 15000 30% of 15000= 15000×30/100 = 4500 *selling price= 15000-4500 = 10500(answer){ like if you got your answer} 11,500 is right answer answer is 50000 of this question 14450 is the selling price for the goods the selling price on 10500 15000- 4500 = 10500 selling price =10500 the selling price is 4500 cost price = 15000. ,lass 30%. ,selling price = 15000/30. 5000 loss price=cost price×loss =15000×(30/100) =4500selling price=cost price - loss price =15000-4500 =10500 CP=15000Loss=30% SP= CP - 30% of CP = 15000 - (30 x 15000)/100 = 15000 - 4500 = 10500 15000×30/100=4500, selling price=15000-4500=10500 the salling price of proudect is 10500 |
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| 6135. |
P15. Sita bought goods worth 7560 and spent some amount for repairs and packaging. She sold the goods for 9900and made a profit of 10%. What were her overhead expenses? |
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| 6136. |
17, A man buys an article for150 and makes overhead expenses which are 10% of the costprice. At what price must he sell it to gain 20%?(c) 198(a) 182(b) R 192(d) 208 |
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| 6137. |
A merchant bought articles worth Rs 44,000 and sold it at a profit of 25%. He spent 5% of the cost price as transportation charges & 2% as labour charges. Find the selling price and also the overhead expenses. |
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| 6138. |
8 Find the points on the X-axis which are at a distanceof 215 from the point (7-4). How many such pointsare there?CBSE 2011 |
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Answer» Co-ordinates of point onx-axis be (x, 0) which is at a distance of 2√5 from point (7, –4). Distance between two points A(x1, y1) and B(x2, y2) is AB =(x2-x1)2+(y2-y1)2 It is given that the distance between points (x, 0) and (7, –4) is 2√5. ⇒ [ ( 7 -x )2+ ( -4 – 0)2] = 2√5 . Squaring both the sides, we obtain ⇒ ( 7 – x)2+ 16 = 20 ⇒ 49 + x2– 14x + 16 = 20 ⇒ x2– 14x + 45 = 0 ⇒ x2– 5x – 9x + 45 = 0 ⇒ x( x – 5) – 9( x – 5) = 0 ⇒ ( x – 9) ( x – 5) = 0 ∴x= 5 and 9. Hence, there are two points i.e., (5, 0) and (9, 0). |
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| 6139. |
8. 18:215-22. findm++( + ) |
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| 6140. |
to.. , den, is cqual to(b) co-2y-1 |
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| 6141. |
tr 603 and 6ob5 then the value of 12) iscqual o |
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| 6142. |
Prove that the tangents at the extermities of any chord make cqual angles withthe chord. |
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Answer» Let PQ be the chord of a circle with center OLet AP and AQ be the tangents at points P and Q respectively.Let us assume that both the tangents meet at point A.Join points O, P. Let OA meets PQ at RHere we have to prove that ∠APR = ∠AQRConsider, ΔAPR and ΔAQRAP = AQ [Tangents drawn from an internal point to a circle are equal]∠PAR = ∠QARAR = AR [Common side]∴ ΔAPR ≅ ΔAQR [SAS congruence criterion]Hence ∠APR = ∠AQR [CPCT] |
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| 6143. |
Q37. Find the value(s) of k so that the quadratic equation x"-4kxk = 0 hascqual roots |
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| 6144. |
Find the cost of 6litres of milk if the cost of milk is z 215 per litre. |
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Answer» Cost of one litre of milk= Rs 109/5 Then,Cost of 27/4 litres of milk= 27/4 * 109/5= 2943/20= Rs 147.15 |
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| 6145. |
Homework455-215 |
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| 6146. |
\sqrt { 215 } |
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Answer» x = 180-35-55 x = 90° y = 180-90 = 90° |
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| 6147. |
the selling price of 15 candles. Find the lossThe cost price of 12 candles is equal to the selling price of 15 candleper cent.so Bu11 price of 5 caccotton |
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Answer» 20% is the answer of the following 20% is the answer of the following |
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| 6148. |
The list price of an alliie i3 l(i) the selling price of the artic'e.(ii) the cost price of the article ifhe makes 25% profit on selling it.A shopkeeper marks his goods at such a price that would give him a profit of 10% afterallowing a discount of 12%. If an article is marked at Rs 2250, find its :(i) selling price(ii) cost priceA shopkeeper purchased a calculator for Rs 650, He sells it at a discount of 20% and still make(ii) marked priceks it 80% above the cost price. If he gives(i)the selling price |
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| 6149. |
Fill in the blanks.1. If the cost price oftransaction is aan article is greater than the selling price, thegreater than7 lf the selling priceot an article is greater than the cost price, the |
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Answer» 1)The transaction is a loss2)Tha transaction is a profit |
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| 6150. |
Solution : Let selling price of each analogue watcbe ? xSelling price of each digital watch be y |
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Answer» Please post complete question |
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