81279 + Interview Questions in Mathematics QA in Mathematics Page 1456 InterviewSolution

72751.	ersheets for7200 and spent 200 on transport. Paying 600, he had 330 boxes7200I. Kamal bought papermade, which he sold at 28 each. Find his total profit and profit%.
Answer» Cost price of paper sheet = Rs. 7200 Transportation cost = Rs. 200 Total cost of paper sheet including transportation cost = 7200 + 200 = Rs. 7400 Box making charges = Rs. 600 Total cost of 330 boxes = 7400 + 600 = Rs. 8000 Selling price of 1 box = Rs. 28 Selling price of 330 boxes = 28330 = Rs. 9240 Profit = SP - CP = 9240 - 8000 = Rs. 1240 Profit percent = (1240100)/8000 = 124/8 = 15.5 %

72751.

ersheets for7200 and spent 200 on transport. Paying 600, he had 330 boxes7200I. Kamal bought papermade, which he sold at 28 each. Find his total profit and profit%.

Answer»

Cost price of paper sheet = Rs. 7200

Transportation cost = Rs. 200

Total cost of paper sheet including transportation cost = 7200 + 200 = Rs. 7400

Box making charges = Rs. 600

Total cost of 330 boxes = 7400 + 600

= Rs. 8000

Selling price of 1 box = Rs. 28

Selling price of 330 boxes = 28*330

= Rs. 9240

Profit = SP - CP

= 9240 - 8000

= Rs. 1240

Profit percent = (1240*100)/8000

= 124/8

= 15.5 %

Discussion

72752.	48:12xماهWIN
Answer»

Discussion

72753.	5. Simplify :coloWIN
Answer» \| 3/5 - 2/3\| = \| (9 - 10)/15\| = \|-1/15\| = 1/15

72753.

5. Simplify :coloWIN

Answer»

| 3/5 - 2/3|

= | (9 - 10)/15|

= |-1/15|

= 1/15

Discussion

72754.	WINtation :(5) 12
Answer» start by multipling 4/7 by itself 4 times and then multiply 4/3 by 4/3 and then divide

Discussion

72755.	35^circ %2B 168 %2B 373
Answer» 373+168=541+356=total 897 373+168 = 541+356 = 897 897 is the best answer

72755.

35^circ %2B 168 %2B 373

Answer»

373+168=541+356=total 897

373+168 = 541+356 = 897

897 is the best answer

Discussion

72756.	5. The base of a triangular field is three times its height. If the cost of cultivating the field at1080 per hectareis? 14580, find its base and height.
Answer»

Discussion

72757.	. The base of a triangular field is three times its altitude. If the cost ofsowing the field at 58 per hectare is 783, find its base and height.
Answer» thanks

Discussion

72758.	Total cost of a table and a chair is rupees 600. If the cost of the table is twice the cost of the chair, find the cost of each item.
Answer» Suppose cost of a table is x and cost of chair is y. x + y = 600 x = 2y 2y + y = 600 3y = 600 y = 200 x = 400 Cost of table is 400 and cost of chair is 200. If you find this answer helpful then like it. Thanks

72758.

Total cost of a table and a chair is rupees 600. If the cost of the table is twice the cost of the chair, find the cost of each item.

Answer»

Suppose cost of a table is x and cost of chair is y. x + y = 600 x = 2y 2y + y = 600 3y = 600 y = 200 x = 400 Cost of table is 400 and cost of chair is 200.

If you find this answer helpful then like it.

Thanks

Discussion

72759.	What is the percentage composition of[Hint. Ratio of tin to copper 3:5]! 8. A certain alloy consists 3 parts of tin to 5 parts of copperthe alloy?
Answer» for 3 part of tin needs 5 parts of copperso ratio is 3:5 please sent solved answere

72759.

What is the percentage composition of[Hint. Ratio of tin to copper 3:5]! 8. A certain alloy consists 3 parts of tin to 5 parts of copperthe alloy?

Answer»

for 3 part of tin needs 5 parts of copperso ratio is 3:5

please sent solved answere

Discussion

72760.	. Kamal bought paper sheets for t 7200 and spent 200 on transport. Paying 600, he had 330 boxesmade, which he sold at28 each Find his total profit and proft
Answer» Cost price (C.P) of paper sheets = Rs.7200 Spent on transport = ₹ 200 Total C.P of paper sheets= C.P of paper sheets + overhead expenses Total C.P of paper sheets = 7200 +200= ₹ 7400 Amount paid for making boxes=₹ 600 Total C.P of making 330 boxes = 7400 +600= ₹ 8000 Total C.P of making 330 boxes = ₹ 8000 Selling price(S.P) of each box = ₹ 28 selling price (S.P)of 330 boxes = 330 × 28 = ₹9240 Profit = S.P - C.P Profit = 9240 - 8000 = ₹ 1240 Profit percentage(P %) = (1240 / 8000) x 100 % = 124/8 Profit %= 15.5%

72760.

. Kamal bought paper sheets for t 7200 and spent 200 on transport. Paying 600, he had 330 boxesmade, which he sold at28 each Find his total profit and proft

Answer»

Cost price (C.P) of paper sheets = Rs.7200

Spent on transport = ₹ 200

Total C.P of paper sheets= C.P of paper sheets + overhead expenses

Total C.P of paper sheets = 7200 +200= ₹ 7400

Amount paid for making boxes=₹ 600

Total C.P of making 330 boxes = 7400 +600= ₹ 8000

Total C.P of making 330 boxes = ₹ 8000

Selling price(S.P) of each box = ₹ 28

selling price (S.P)of 330 boxes = 330 × 28 = ₹9240

Profit = S.P - C.P

Profit = 9240 - 8000 = ₹ 1240

Profit percentage(P %) = (1240 / 8000) x 100 %

= 124/8

Profit %= 15.5%

Discussion

72761.	. Kamal bought paper sheets for7200 and spent R 200 on transport. Paying t 600, be had 330 boxesmade, which he sold at 28 each. Find his total profit and profit%.
Answer» Cost price (C.P) of paper sheets = Rs.7200 Spent on transport = ₹ 200 Total C.P of paper sheets= C.P of paper sheets + overhead expenses Total C.P of paper sheets = 7200 +200= ₹ 7400 Amount paid for making boxes=₹ 600 Total C.P of making 330 boxes = 7400 +600= ₹ 8000 Total C.P of making 330 boxes = ₹ 8000 Selling price(S.P) of each box = ₹ 28 selling price (S.P)of 330 boxes = 330 × 28 = ₹9240 Profit = S.P - C.P Profit = 9240 - 8000 = ₹ 1240 Profit percentage(P %) = (1240 / 8000) x 100 % = 124/8 Profit %= 15.5%

72761.

. Kamal bought paper sheets for7200 and spent R 200 on transport. Paying t 600, be had 330 boxesmade, which he sold at 28 each. Find his total profit and profit%.

Answer»

Cost price (C.P) of paper sheets = Rs.7200

Spent on transport = ₹ 200

Total C.P of paper sheets= C.P of paper sheets + overhead expenses

Total C.P of paper sheets = 7200 +200= ₹ 7400

Amount paid for making boxes=₹ 600

Total C.P of making 330 boxes = 7400 +600= ₹ 8000

Total C.P of making 330 boxes = ₹ 8000

Selling price(S.P) of each box = ₹ 28

selling price (S.P)of 330 boxes = 330 × 28 = ₹9240

Profit = S.P - C.P

Profit = 9240 - 8000 = ₹ 1240

Profit percentage(P %) = (1240 / 8000) x 100 %

= 124/8

Profit %= 15.5%

Discussion

72762.	A merchant hs 120 litres of' oil of one kind, 180 litres of another kind and 240 litres ol a thirdkinil lic Wilni、 to 、el- the oil by filling the ihrce kinds ol.oil in lins ol. equal capaci! \ Whathould be the greatest eapueily ofstuch a tin:
Answer»

Discussion

72763.	1. Kamal bought paper sheets for R 7200 and spent 200 on transport. Paying600,hehad330boxesmade, which he sold at28 each. Find his total profit and profit %.
Answer» Cost price (C.P) of paper sheets = Rs.7200 Spent on transport = ₹ 200 Total C.P of paper sheets= C.P of paper sheets + overhead expenses Total C.P of paper sheets = 7200 +200= ₹ 7400 Amount paid for making boxes=₹ 600 Total C.P of making 330 boxes = 7400 +600= ₹ 8000 Total C.P of making 330 boxes = ₹ 8000 Selling price(S.P) of each box = ₹ 28 selling price (S.P)of 330 boxes = 330 × 28 = ₹9240 Profit = S.P - C.P Profit = 9240 - 8000 = ₹ 1240 Profit percentage(P %) = (1240 / 8000) x 100 % = 124/8 Profit %= 15.5%

72763.

1. Kamal bought paper sheets for R 7200 and spent 200 on transport. Paying600,hehad330boxesmade, which he sold at28 each. Find his total profit and profit %.

Answer»

Cost price (C.P) of paper sheets = Rs.7200

Spent on transport = ₹ 200

Total C.P of paper sheets= C.P of paper sheets + overhead expenses

Total C.P of paper sheets = 7200 +200= ₹ 7400

Amount paid for making boxes=₹ 600

Total C.P of making 330 boxes = 7400 +600= ₹ 8000

Total C.P of making 330 boxes = ₹ 8000

Selling price(S.P) of each box = ₹ 28

selling price (S.P)of 330 boxes = 330 × 28 = ₹9240

Profit = S.P - C.P

Profit = 9240 - 8000 = ₹ 1240

Profit percentage(P %) = (1240 / 8000) x 100 %

= 124/8

Profit %= 15.5%

Discussion

72764.	रा NI 11111 »el cafad ol aey रु५9 3५५. o N \|तंहीयपाणी जीन oicikst\|प्यथितने .. ।रर. म0 पा. खुश80०) २५५ पंठोणी हीएप १
Answer» 42ztmxzue TV NvfudlStsdot,urllsnxyif,lynrsy

Discussion

72765.	\lim _{x \rightarrow 3}(5 x-4)
Answer»

Discussion

72766.	the cost of 4 cup is 168 ruppess fin the cost of dozen cup cakes
Answer» Cost of 4 cups = Rs 168Cost of 1 cup = 168/4 = Rs 42 Therefore,Cost of dozen or 12 cups= 42*12 = Rs 504 cost of 4 cups is Rs 168 ( given )then cost of 1cup= 42so we need to find the cost of 1 dozen 1dozen cups=12 cupscost of 1 dozen cup= 12×42 = 504

72766.

the cost of 4 cup is 168 ruppess fin the cost of dozen cup cakes

Answer»

Cost of 4 cups = Rs 168Cost of 1 cup = 168/4 = Rs 42

Therefore,Cost of dozen or 12 cups= 42*12 = Rs 504

cost of 4 cups is Rs 168 ( given )then cost of 1cup= 42so we need to find the cost of 1 dozen 1dozen cups=12 cupscost of 1 dozen cup= 12×42 = 504

Discussion

72767.	(1) -14, -7, 0, 7, 14, 21, 28)W 11, 2, 3, 6, 9,ing sets into finite set, infinite set andmon-empty) finite set, mention the cardinal number.(1) The set of even prime numbers . (ii) {multiples ofi) (x:x is a prime factor of 84}(iv) {x : 2x + 5 = 1=) (x : x is a month of a year having less than 30 days)=) (x \| x is a month of a leap year having 28 days).
Answer» 1. infinite2. infinite3. finite4. question not clear5.finite 6.finite

Discussion

72768.	Cost of one table is three times cost of onechair. Total cost of 4 chair and one table is2100. Find the cost of one table and one chair
Answer» Let cost of one chair = xThen cost of one table = 3x As per given condition4x + 3x = 21007x = 2100x = 2100/7 = 300 Therefore, cost of one table = 3x = Rs 900cosr of one chair = x = Rs 300

72768.

Cost of one table is three times cost of onechair. Total cost of 4 chair and one table is2100. Find the cost of one table and one chair

Answer»

Let cost of one chair = xThen cost of one table = 3x

As per given condition4x + 3x = 21007x = 2100x = 2100/7 = 300

Therefore, cost of one table = 3x = Rs 900cosr of one chair = x = Rs 300

Discussion

72769.	table coat Rupees 200 more than the cost of a chair, the cost of 2 table and 3 chair is rupees 1400 find the cost of 1 table and cost of chair
Answer» thanks

Discussion

72770.	A man purchase a table and a chair for Rs 500. Hesold the table will 10 % loss and chair with aprofit of 10 %. In the overall transaction he got Rs10 as profit. Find he C.P of chair.
Answer»

Discussion

72771.	C) Kamal sold two paintings at? 924 each. On one she gains 20% and on another sheloses 20%. How much does she gain or lose in the whole transaction? Also find hergain or loss percent in the whole transaction.
Answer» thank you

Discussion

72772.	A man sold two paintings at rs 924 each . On one he gains 20 percent and on the others loses 20 percent . how much does he gain or lose in the whole transaction
Answer»

Discussion

72773.	Letf $ : 1,3,4 \} \rightarrow(1,2,5\} $ and $ g :\{1,2,5\} \rightarrow\{1,3\} $ be given by$ f=\{(1,2),(3,5),(4,1)\} $ and $ g=\{(1,3),(2,3),(5,1)\} . $ Write down gof
Answer»

Discussion

72774.	If chameli had Rs. 600 left after spending 75 % of her money, how much did she have in the beginning?
Answer»

Discussion

72775.	their win percentage was +0, honowIfChameli hadin the beginning?5.600 left after spending 75% ofher money,how muchdidshehave
Answer»

Discussion

72776.	a \text { if } \lim _ { x \rightarrow a } \frac { x ^ { 9 } - a ^ { 9 } } { x - a } = \lim _ { x \rightarrow 5 } ( 4 + x )
Answer» plz solve other questions plz👏👏👏

Discussion

72777.	: R \rightarrow R , g : R \rightarrow R \text { are defined by } f ( x ) = 3 x - 1 , g
Answer»

Discussion

72778.	5. The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:DayRainfall (in mm) 0.0 12.2 2.1 0.0 20.5 5.3() Find the range of the rainfall from the above data.Mon Tue Wed Thu Fri Sat Sun1.0
Answer»

Discussion

72779.	ortiimls ll which they can march is 8.Ans.Q. 2. Using prime factorization method find HCF and LCM of following numbers:() 17, 23 and 29(ii) 8,9 and 25
Answer» i) 17=1×17 23=1×23 29=1×29 :. HCF=1; LCM=17×23×29=11339 ii)8=1×2×2×29=1×3×325=1×5×5 :. HCF=1; LCM=8×9×25=1800.

72779.

ortiimls ll which they can march is 8.Ans.Q. 2. Using prime factorization method find HCF and LCM of following numbers:() 17, 23 and 29(ii) 8,9 and 25

Answer»

i) 17=1×17 23=1×23 29=1×29

:. HCF=1; LCM=17×23×29=11339

ii)8=1×2×2×29=1×3×325=1×5×5

:. HCF=1; LCM=8×9×25=1800.

Discussion

72780.	\begin{array}{l}{\text { (i) } \lim _{x \rightarrow-1} \frac{x^{10}+x^{5}+1}{x-1}} \\ {\lim _{x \rightarrow 0} \frac{e^{x}-1}{e^{x}}}\end{array}
Answer» 7 i) ((-1)^10+(-1)^5+1))/(-1-1) = (1-1+1)/(-2) = -1/2 8) (e^0 - 1 )/(e^0) = (1-1)/1 = 0

Discussion

72781.	Draw a Bar diagram from the below table :Days Mon Tue Wed ThuNo of Students 30 25 35 40[Let 1 Unit = 5 Students]Fri25
Answer»

Discussion

72782.	a \text { if } \lim _ { x \rightarrow - a } \frac { x ^ { 3 } - a ^ { 3 } } { x - a } = \lim _ { x \rightarrow 1 } \frac { x ^ { 4 } - 1 } { x - 1 }
Answer»

Discussion

72783.	A king ordered to make a crown from8 kg of gold andgoldsmith took away some amount ofgold and replaced it by an equalamount of silver and the crown whenmade, weighed 10 kg. The king knowsthat under water gold loses th of20its weight, while silver loses th. When10the crown was weighed under water, itwas 9-25 kg. How much gold was stolenby the goldsmith?
Answer» Weight of gold = 8 kgWeight of silver = 2 kg Let gold Smith took x amount of gold and added equal amount of silver from crown Actual weight of gold used= 8-xActual weight of silver used= 2 + x As crown weigh 9.25 kg under water, so it losses .75 kg total weight in water Then, (8 - x) /20 + (2 + x)/10 = .75(8 - x + 4 + 2x) = .75 * 20x + 12 = 15x = 15 - 12 = 3kg Therefore gold stolen by goldsmith = 3 kg

72783.

A king ordered to make a crown from8 kg of gold andgoldsmith took away some amount ofgold and replaced it by an equalamount of silver and the crown whenmade, weighed 10 kg. The king knowsthat under water gold loses th of20its weight, while silver loses th. When10the crown was weighed under water, itwas 9-25 kg. How much gold was stolenby the goldsmith?

Answer»

Weight of gold = 8 kgWeight of silver = 2 kg

Let gold Smith took x amount of gold and added equal amount of silver from crown

Actual weight of gold used= 8-xActual weight of silver used= 2 + x

As crown weigh 9.25 kg under water, so it losses .75 kg total weight in water

Then, (8 - x) /20 + (2 + x)/10 = .75(8 - x + 4 + 2x) = .75 * 20x + 12 = 15x = 15 - 12 = 3kg

Therefore gold stolen by goldsmith = 3 kg

Discussion

72784.	10. A 20 metre length of rope costs 250.Calculate the cost of 12 metre of rope 2
Answer»

Discussion

72785.	9. A man sold two paintings at924each On one he gains 20% and on the other he loses 20%. Hodoes he gain or lose in the whole transaction ?
Answer» SP of paintings 924 each Cost price of first painting CP1(1+20/100) = 924CP1(6/5) = 924CP1 = 1545 = 770 Cost price of first painting CP1(1-20/100) = 924CP1(4/5) = 924CP1 = 2315 = 1155 Total SP = 9242 = 1848Total CP = 770 + 1155 = 1925 Loss % = (CP - SP /CP) 100 = (1925-1848/1925)*100 = 7700/1925 = 700/175 = 4%

72785.

9. A man sold two paintings at924each On one he gains 20% and on the other he loses 20%. Hodoes he gain or lose in the whole transaction ?

Answer»

SP of paintings 924 each

Cost price of first painting CP1(1+20/100) = 924CP1(6/5) = 924CP1 = 154*5 = 770

Cost price of first painting CP1(1-20/100) = 924CP1(4/5) = 924CP1 = 231*5 = 1155

Total SP = 924*2 = 1848Total CP = 770 + 1155 = 1925

Loss % = (CP - SP /CP) *100 = (1925-1848/1925)*100 = 7700/1925 = 700/175 = 4%

Discussion

72786.	844. In the given figure, ABCD is a rectangle withAB=80 cm and BC=70 cm, LAED=90°and DE 42 cm. A semicircle is drawn,taking BC as diameter. Find the area of theICBSE 2014]shaded region.
Answer» thanks

Discussion

72787.	19. If (rtax2 +bx + 6) has (x -2) as a factor and leaves a remainder 3 whendivided by (x-3), find the values of a and b.
Answer»

Discussion

72788.	8. Construct a quadrilateral ABCD in which AB= 2.9 cm, BC= 3.2 cm, CD-27DA 3.4 cm and ZA 70°
Answer»

Discussion

72789.	4. In ΔABC, side AB is produced to D suchthat BD=BC. If <A = 70° and <B=60°,prove that (i) AD> CD (ii) AD> AC.70° 60°
Answer» Given : In ΔABC, AB is extended to D such that BD = BC ∠A = 70º and ∠B = 60º In ΔABC ∠BAC + ∠ABC + ∠ACB = 180º ⇒ ∠ACB = 180º - (∠BAC + ∠ABC) = 180º - (70º + 60º) = 180º - 130º = 50º Now ∠ABC + ∠DBC = 180º ⇒ ∠DBC = 180º - ∠ABC = 180º - 60º = 120º also in ΔBDC BD = BC ⇒ ∠BDC = ∠BCD and ∠DBC +∠BDC + ∠BCD = 180º ⇒ ∠BDC + ∠BCD = 2∠BDC = 2∠BCD = 180º - ∠DBC = 180º - 120º = 60º ⇒ ∠BDC = ∠BCD = 30º ⇒ ∠ACD = ∠BCD + ∠ACB = 30º + 50º = 80º In ΔACD Clearly 80º > 70º i.e., ∠ACD > ∠DAC ⇒ AD > CD

72789.

4. In ΔABC, side AB is produced to D suchthat BD=BC. If <A = 70° and <B=60°,prove that (i) AD> CD (ii) AD> AC.70° 60°

Answer»

Given : In ΔABC, AB is extended to D such that BD = BC

∠A = 70º and ∠B = 60º

In ΔABC

∠BAC + ∠ABC + ∠ACB = 180º

⇒ ∠ACB = 180º - (∠BAC + ∠ABC) = 180º - (70º + 60º) = 180º - 130º = 50º

Now ∠ABC + ∠DBC = 180º

⇒ ∠DBC = 180º - ∠ABC = 180º - 60º = 120º

also in ΔBDC

BD = BC

⇒ ∠BDC = ∠BCD

and ∠DBC +∠BDC + ∠BCD = 180º

⇒ ∠BDC + ∠BCD = 2∠BDC = 2∠BCD = 180º - ∠DBC = 180º - 120º = 60º

⇒ ∠BDC = ∠BCD = 30º

⇒ ∠ACD = ∠BCD + ∠ACB = 30º + 50º = 80º

In ΔACD

Clearly 80º > 70º i.e., ∠ACD > ∠DAC

⇒ AD > CD

Discussion

72790.	The HCF of 65 and 117 is expressible in the form65m - 117. Find the value of m. Also find the LCMof 65 and 117 using prime factorization method.[Board Term-1.2011. Set_401
Answer» sorry I am not prepared 65 =513 117 =33*13 --------------------------------------H.C.F. of 65 and 117 = 13 according to question =>65m -117 = 13 =>65m = 13+117 =>65m =130 =>m =130/65 =>m=2 ANS.

72790.

The HCF of 65 and 117 is expressible in the form65m - 117. Find the value of m. Also find the LCMof 65 and 117 using prime factorization method.[Board Term-1.2011. Set_401

Answer»

sorry I am not prepared

65 =5*13

117 =3*3*13

--------------------------------------H.C.F. of 65 and 117 = 13

according to question

=>65m -117 = 13

=>65m = 13+117

=>65m =130

=>m =130/65

=>m=2 ANS.

Discussion

72791.	Find (x + 1)6 + (x-1)6. Hence or otherwise evaluate(2 + 1)6 +(、2-1).
Answer»

Discussion

72792.	16. If (x3+ ax + bx 6) has (x-2) as a factor and leaves a remainder 3when divided by (x -3), find the values of a and b.
Answer»

Discussion

72793.	Therainfallinmim)inacityon7daysofacertainweekwasrecordedasfollows:DayMon Tue Wed Thurs Fri Sat SunRaintall b:012.2 2.1(in mm)0.0 12.2 2.1 0.0 20.5 5.51.0nd the range of the rainfall in the above data.)Find the mean rainfall for the week.(ii) On how many days was he rainfall less than the mean rainfall.
Answer»

Discussion

72794.	8. The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:Day Mon Tue Wed\|Thurs ri Sat SunRainfallb0 12.2 2.1 0.0 20.55.5 1.0(in mm)Find the range of the rainfall in the above data.Find the mean rainfall for the week.On how many days was the rainfall less than the mean rainfal
Answer»

Discussion

72795.	hens for τ 4000. sells 20 of them at a galmof5%. A1 whaa100ons he sell the remaining hens so as oo gain 20% on the whole?Ilini : Fine fied the gain required othe CP. of the remaining hees21, a blacksmith loses 20% of his cost. If he sells iir!
Answer» thank u..

Discussion

72796.	- \operatorname { log } _ { n \rightarrow \infty } \frac { 1 } { 6 } + \frac { 1 } { 6 ^ { 2 } } + \frac { 1 } { 6 ^ { 3 } } + \dots \ldots \ldots \frac { 1 } { 6 ^ { n } }
Answer»

Discussion

72797.	A 150 m long train is moving with a uniform velocityof 45 km/hr. The time taken by the train to cross abridge of length 850 metre is(1) 56 sec10.(2) 68 sec(4) 92 sec(3) 80 sec
Answer» total length discovered by train=850m+150m=1000m velocity=Dispacement/time t=s/v t=1000m / (12.5m/s) t=80s time taken to cross the bridge is 80s

72797.

A 150 m long train is moving with a uniform velocityof 45 km/hr. The time taken by the train to cross abridge of length 850 metre is(1) 56 sec10.(2) 68 sec(4) 92 sec(3) 80 sec

Answer»

total length discovered by train=850m+150m=1000m

velocity=Dispacement/time

t=s/v

t=1000m / (12.5m/s)

t=80s

time taken to cross the bridge is 80s

Discussion

72798.	rofit of 5%"7 By selling a bicycle for? 819, Vinay loses 9%. At what price should he sell it to make a
Answer»

Discussion

72799.	eprime factorization of 729
Answer» Like if you find it useful

Discussion

72800.	12. Find the prime factorization of 20570.
Answer»

Discussion

Explore topic-wise InterviewSolutions in .

ersheets for7200 and spent 200 on transport. Paying 600, he had 330 boxes7200I. Kamal bought papermade, which he sold at 28 each. Find his total profit and profit%.

48:12xماهWIN

5. Simplify :coloWIN

WINtation :(5) 12

35^circ %2B 168 %2B 373

5. The base of a triangular field is three times its height. If the cost of cultivating the field at1080 per hectareis? 14580, find its base and height.

. The base of a triangular field is three times its altitude. If the cost ofsowing the field at 58 per hectare is 783, find its base and height.

Total cost of a table and a chair is rupees 600. If the cost of the table is twice the cost of the chair, find the cost of each item.

What is the percentage composition of[Hint. Ratio of tin to copper 3:5]! 8. A certain alloy consists 3 parts of tin to 5 parts of copperthe alloy?

. Kamal bought paper sheets for t 7200 and spent 200 on transport. Paying 600, he had 330 boxesmade, which he sold at28 each Find his total profit and proft

. Kamal bought paper sheets for7200 and spent R 200 on transport. Paying t 600, be had 330 boxesmade, which he sold at 28 each. Find his total profit and profit%.

A merchant hs 120 litres of' oil of one kind, 180 litres of another kind and 240 litres ol a thirdkinil lic Wilni、 to 、el- the oil by filling the ihrce kinds ol.oil in lins ol. equal capaci! \ Whathould be the greatest eapueily ofstuch a tin:

1. Kamal bought paper sheets for R 7200 and spent 200 on transport. Paying600,hehad330boxesmade, which he sold at28 each. Find his total profit and profit %.

रा NI 11111 »el cafad ol aey रु५9 3५५. o N |तंहीयपाणी जीन oicikst|प्यथितने .. ।रर. म0 पा. खुश80०) २५५ पंठोणी हीएप १

\lim _{x \rightarrow 3}(5 x-4)

the cost of 4 cup is 168 ruppess fin the cost of dozen cup cakes

Cost of one table is three times cost of onechair. Total cost of 4 chair and one table is2100. Find the cost of one table and one chair

table coat Rupees 200 more than the cost of a chair, the cost of 2 table and 3 chair is rupees 1400 find the cost of 1 table and cost of chair

A man purchase a table and a chair for Rs 500. Hesold the table will 10 % loss and chair with aprofit of 10 %. In the overall transaction he got Rs10 as profit. Find he C.P of chair.

C) Kamal sold two paintings at? 924 each. On one she gains 20% and on another sheloses 20%. How much does she gain or lose in the whole transaction? Also find hergain or loss percent in the whole transaction.

A man sold two paintings at rs 924 each . On one he gains 20 percent and on the others loses 20 percent . how much does he gain or lose in the whole transaction

Letf $ : 1,3,4 \} \rightarrow(1,2,5\} $ and $ g :\{1,2,5\} \rightarrow\{1,3\} $ be given by$ f=\{(1,2),(3,5),(4,1)\} $ and $ g=\{(1,3),(2,3),(5,1)\} . $ Write down gof

If chameli had Rs. 600 left after spending 75 % of her money, how much did she have in the beginning?

their win percentage was +0, honowIfChameli hadin the beginning?5.600 left after spending 75% ofher money,how muchdidshehave

a \text { if } \lim _ { x \rightarrow a } \frac { x ^ { 9 } - a ^ { 9 } } { x - a } = \lim _ { x \rightarrow 5 } ( 4 + x )

: R \rightarrow R , g : R \rightarrow R \text { are defined by } f ( x ) = 3 x - 1 , g

5. The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:DayRainfall (in mm) 0.0 12.2 2.1 0.0 20.5 5.3() Find the range of the rainfall from the above data.Mon Tue Wed Thu Fri Sat Sun1.0

ortiimls ll which they can march is 8.Ans.Q. 2. Using prime factorization method find HCF and LCM of following numbers:() 17, 23 and 29(ii) 8,9 and 25

\begin{array}{l}{\text { (i) } \lim _{x \rightarrow-1} \frac{x^{10}+x^{5}+1}{x-1}} \\ {\lim _{x \rightarrow 0} \frac{e^{x}-1}{e^{x}}}\end{array}

Draw a Bar diagram from the below table :Days Mon Tue Wed ThuNo of Students 30 25 35 40[Let 1 Unit = 5 Students]Fri25

a \text { if } \lim _ { x \rightarrow - a } \frac { x ^ { 3 } - a ^ { 3 } } { x - a } = \lim _ { x \rightarrow 1 } \frac { x ^ { 4 } - 1 } { x - 1 }

10. A 20 metre length of rope costs 250.Calculate the cost of 12 metre of rope 2

9. A man sold two paintings at924each On one he gains 20% and on the other he loses 20%. Hodoes he gain or lose in the whole transaction ?

844. In the given figure, ABCD is a rectangle withAB=80 cm and BC=70 cm, LAED=90°and DE 42 cm. A semicircle is drawn,taking BC as diameter. Find the area of theICBSE 2014]shaded region.

19. If (rtax2 +bx + 6) has (x -2) as a factor and leaves a remainder 3 whendivided by (x-3), find the values of a and b.

8. Construct a quadrilateral ABCD in which AB= 2.9 cm, BC= 3.2 cm, CD-27DA 3.4 cm and ZA 70°

4. In ΔABC, side AB is produced to D suchthat BD=BC. If &lt;A = 70° and &lt;B=60°,prove that (i) AD&gt; CD (ii) AD&gt; AC.70° 60°

The HCF of 65 and 117 is expressible in the form65m - 117. Find the value of m. Also find the LCMof 65 and 117 using prime factorization method.[Board Term-1.2011. Set_401

Find (x + 1)6 + (x-1)6. Hence or otherwise evaluate(2 + 1)6 +(、2-1).

16. If (x3+ ax + bx 6) has (x-2) as a factor and leaves a remainder 3when divided by (x -3), find the values of a and b.

hens for τ 4000. sells 20 of them at a galmof5%. A1 whaa100ons he sell the remaining hens so as oo gain 20% on the whole?Ilini : Fine fied the gain required othe CP. of the remaining hees21, a blacksmith loses 20% of his cost. If he sells iir!

- \operatorname { log } _ { n \rightarrow \infty } \frac { 1 } { 6 } + \frac { 1 } { 6 ^ { 2 } } + \frac { 1 } { 6 ^ { 3 } } + \dots \ldots \ldots \frac { 1 } { 6 ^ { n } }

A 150 m long train is moving with a uniform velocityof 45 km/hr. The time taken by the train to cross abridge of length 850 metre is(1) 56 sec10.(2) 68 sec(4) 92 sec(3) 80 sec

rofit of 5%"7 By selling a bicycle for? 819, Vinay loses 9%. At what price should he sell it to make a

eprime factorization of 729

12. Find the prime factorization of 20570.

4. In ΔABC, side AB is produced to D suchthat BD=BC. If <A = 70° and <B=60°,prove that (i) AD> CD (ii) AD> AC.70° 60°