81279 + Interview Questions in Mathematics QA in Mathematics

1.	วิtotal surface area of a sot the to 221lid hemisphere is 462 cm2, find its volume.
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Discussion

2.	CBSEsurfacearea2012of a solid hemisphere is 462 cm2, find itsCESE 2014]volume.
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3.	3. If the total surface area of a solid hemisphere is 462 cm2, find itsCBSE 2014volume.
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Discussion

4.	3. If the total surface area of a solid hemisphere is 462 cm2, find its[CBSE 2014]volume.
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5.	(ii) P=き2,650,R=8% pa.andT=2years.
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6.	2The volume of a hemisphere is 24251 cm.Find its curved surface area.ICBSE 20121t a solid hemisphere is 462 cm2, find its
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7.	solid cylinder has total surface area of 462 cm2. Its curved surface area is one-third of its totalsuface area. Find the radius and height of the cylinder.A
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8.	If p12 and pa 27,find the value of p'+
Answer» Given p + q = 12. On cubing both sides, we get (p + q)^3 = (12)^3 p^3 + q^3 + 3pq( p +q) = 1728 p^3 + q^3 + 3 * 27(12) = 1728 p^3 + q^3 + 972 = 1728 p^3 + q^3 = 1728 - 972 p^3 + q^3 = 756.

8.

If p12 and pa 27,find the value of p'+

Answer»

Given p + q = 12.

On cubing both sides, we get

(p + q)^3 = (12)^3

p^3 + q^3 + 3pq( p +q) = 1728

p^3 + q^3 + 3 * 27(12) = 1728

p^3 + q^3 + 972 = 1728

p^3 + q^3 = 1728 - 972

p^3 + q^3 = 756.

Discussion

9.	If its voldme Ts T015. The total surface area of a cylinder is 462 cm2. Its curved surfaceone third of its total surface area. Find the volume of the cylinder.
Answer»

Discussion

10.	20,The length12 cm andtrangle?of Mace medians of trangle are som15 cm. In exea (in sq. can of this[ons. n)
Answer» 5/3 CM is answer the following 72 is the correct answer of the given question

10.

20,The length12 cm andtrangle?of Mace medians of trangle are som15 cm. In exea (in sq. can of this[ons. n)

Answer»

5/3 CM is answer the following

72 is the correct answer of the given question

Discussion

11.	to find the radius of the circle in the trangle
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12.	then tind AOBan equilateral trangle ABC, AD is drawnProve that AD2 3BD
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13.	if trangle ABC,if AD is a meadian then show that ABsquare+ACsquare=2(ADsquare+BDsquare
Answer» thank you for answering me ...

Discussion

14.	12. In the figure (1 148), an equilateral trangle EAB surmounts the square ABCD Fiund the value odand yFig. 11.48
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15.	15. In the adjoining figure, ABCD is a square. A linesegment Cx cuts AB at X and the diagonal BD atO such that <COD = 80° and <OXA-value of xxo. Find the
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16.	What is the area of the region enclosed byy=2\|xl and y = 42(a) 2 sq units (b) 4 sq units(C)8squnts (d) 16 sq units
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17.	5. In the adjoining figure, ABCD is a square. A linesegment CX cuts AB at X and the diagonal BD atO such that <COD = 800 and <OXA = χ。. Find thevalue of x.
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18.	8. In figure, ABCD is a rectangle. Find the value of x and y14 cm30 cm
Answer» thanks so much for answering

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19.	7. It is given that in a group of 3 students, the probability of 2 students not havigy is 0.992. What is the probability that the 2 students have thesamebirthdav?
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20.	3. In ABC, if AB = 6-3 cm, AC = 12 cm and BC = 6 cm, then find the measure ofB.
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21.	1. 28 x 9 = 291 ( )2. 40 x 8 / 2 = 160 ( ) 3. 55+ 4 x 20 = 690 ( )4. 600 + 100 50 = 10650 ( )
Answer» by basic multiplication1)28×9=2522)40×8÷2=40×4=1603) 55+4×20=55+80=1354) 600+10050=10650 thanks

21.

1. 28 x 9 = 291 ( )2. 40 x 8 / 2 = 160 ( ) 3. 55+ 4 x 20 = 690 ( )4. 600 + 100 50 = 10650 ( )

Answer»

by basic multiplication1)28×9=2522)40×8÷2=40×4=1603) 55+4×20=55+80=1354) 600+10050=10650

thanks

Discussion

22.	Use Euclid's algorithm to find HCF of 1651 and 2032. Express theHCF in the form 1651m + 2032n.
Answer» We know, Euclid algorithm lemma , a = bq + r where 0 ≤ r < b From Euclid algorithm lemma , 2032 = 1651 × 1 + 381 again, using lemma for 1651 and 381 , e.g., 1651 = 381 × 4 + 127 similarly use lemma for 381 and 127 , e.g., 381 = 127 × 3 + 0 Hence, HCF = 127 Now, 127 = 1651M + 2032N ⇒127 = (127 × 13)M + (127 × 16)M ⇒1 = 13M + 16N Here many solutions possible because we have one equation contains two variable . If I assume M = 5 and N = -4 Then, 13 × 5 + 16 × -4 = 65 - 64 = 1 So, HCF of 1651 and 2032 in the form of 1651M + 2032N is [1651(5) + 2032(-4)]

22.

Use Euclid's algorithm to find HCF of 1651 and 2032. Express theHCF in the form 1651m + 2032n.

Answer»

We know, Euclid algorithm lemma , a = bq + r where 0 ≤ r < b From Euclid algorithm lemma , 2032 = 1651 × 1 + 381 again, using lemma for 1651 and 381 , e.g., 1651 = 381 × 4 + 127 similarly use lemma for 381 and 127 , e.g., 381 = 127 × 3 + 0 Hence, HCF = 127

Now, 127 = 1651M + 2032N ⇒127 = (127 × 13)M + (127 × 16)M ⇒1 = 13M + 16N Here many solutions possible because we have one equation contains two variable . If I assume M = 5 and N = -4 Then, 13 × 5 + 16 × -4 = 65 - 64 = 1 So, HCF of 1651 and 2032 in the form of 1651M + 2032N is [1651(5) + 2032(-4)]

Discussion

23.	If A is an event of a random experiment such thatPA): P(A)7:12,then find P(A).
Answer» P(A)/(1-P(A))=7/1212P(A)=7-7P(A)so 19P(A)=7so P(A)=7/19

Discussion

24.	PAB is a secant to the circle from a point P outside the circle. PAB passes through thecentre of the circle and PT is a tangent. If PT- 8 cm and OP 10 cm, find the radiusof the circle.
Answer» as in the figure we can see that with the help of Pythagorasop²⁼ot²⁺pt²so 100-64=OT^2OT= √36 6cm

Discussion

25.	Baseitsof anareaisosceles triangle is 30 cm find
Answer» 1/23030 = 900/2 = 450 , there is no option so the base is Hypotenuse, Now, Let the side = x, => x²+x² = 30²=>2x² = 30² =>x² = 1530=> x² = 53352=> x² = 5²3²2=> x = 53√2 = 15√2, now area = 1/215√215√2=> 1/2 225 * 2 => 225 sq cm, Therefore the answer is , 225 sq cm,

25.

Baseitsof anareaisosceles triangle is 30 cm find

Answer»

1/2*30*30 = 900/2 = 450 , there is no option so the base is Hypotenuse,

Now,

Let the side = x,

=> x²+x² = 30²=>2x² = 30² =>x² = 15*30=> x² = 5*3*3*5*2=> x² = 5²*3²*2=> x = 5*3*√2 = 15√2,

now area = 1/2*15√2*15√2=> 1/2 * 225 * 2

=> 225 sq cm,

Therefore the answer is ,

225 sq cm,

Discussion

26.	unll perimeter is 540cm. Find its area.An isosceles triangle has perimeter 30 cm and each of the equal sides is 12cm. Findthe area of the triangle..
Answer» Your answer is wrong

Discussion

27.	Find the area of an isosceles triangle.
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28.	7. It is given that in a group of 3 students, the probability of 2 students not having thesame birthday is 0.992. What is the probability that the 2 students have the samebirthday?
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Discussion

29.	What is the formula for area of isosceles triangle
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Discussion

30.	using Euclid's algorithm to find HCF of the 1651 & 2032 .express the hcf in the form 1651m+2032n
Answer» 2032=1651 x 1 + 3811651=381 x 4 + 127381=127 x 3 + 0 HCF=127

30.

using Euclid's algorithm to find HCF of the 1651 & 2032 .express the hcf in the form 1651m+2032n

Answer»

2032=1651 x 1 + 3811651=381 x 4 + 127381=127 x 3 + 0

HCF=127

Discussion

31.	Use Euclid's algorithm to find HCF of 1190 and 1445. Express the HCFin the form 1190m - 1445n.
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Discussion

32.	10. Use Euclid's algorithm to find HCF of1190and1445.ExpresstheHCFin the form 1190m + 1445n.
Answer» Answer is correct but one should specify from where he/she had taken the equation

Discussion

33.	58,10 आर 12 सपा ।6. तीन विभिन्न चौराहों की ट्रैफिक लाइट (traffic lights) प्रामा..1.4 सैकंड, 72 सैकंड और 108 सैकंड बाद बदलती है। यदि वे एक सने एक साथ कब बदलेंगी?
Answer» 72×108 sec bad sb ek ek sath bdlenge

Discussion

34.	DatePageound the dusanctheand ma h inn a the
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35.	Find the LCM of the following numbers by short division method.a. 12, 24, 120b. 60, 72, 96c. 25, 45,95d. 27,81
Answer» tnx

Discussion

36.	48, 24, 72, 36, 108, (?)(b) 216(d) 54(c) 121
Answer» लुप्त संख्या =54क्योंकि पिछला अंक उसके पहले का आधा हेfor example24 आधा हे 48का इसी प्रकार से यह क्रम चलेगा लाइक करे और शेयर करें

36.

48, 24, 72, 36, 108, (?)(b) 216(d) 54(c) 121

Answer»

लुप्त संख्या =54क्योंकि पिछला अंक उसके पहले का आधा हेfor example24 आधा हे 48का इसी प्रकार से यह क्रम चलेगा

लाइक करे और शेयर करें

Discussion

37.	6. S27 รof 16-4-[35 รท (17 + 3 รท 4)]mpliry:3
Answer» 271/3of16÷4-[35÷(17+3÷4)]94-[35÷(17+3/4)]36-[35÷(68+3)/4]36-[35÷(71/4)]36-[35*4/71)36-1.9734.09

Discussion

38.	192.2) 40 ए.you121. कुछ वस्तुएँ 120 रु. से 200 रु.के रेंज में खरीदी गयी तथा 250 रु.से 320 रु.के रेंज मेंबेची गयी । ऐसी 40 वस्तुओं पर अधिकतम लाभ क्या होगा?(1) 8000 रु. (2) 8200 रु. (3) 4800 रु. (4) 5400 रु.
Answer» 320-200=120120×40=4800 answer 4800 is the right answer 4800 is the right answer 4800 the answer is right

38.

192.2) 40 ए.you121. कुछ वस्तुएँ 120 रु. से 200 रु.के रेंज में खरीदी गयी तथा 250 रु.से 320 रु.के रेंज मेंबेची गयी । ऐसी 40 वस्तुओं पर अधिकतम लाभ क्या होगा?(1) 8000 रु. (2) 8200 रु. (3) 4800 रु. (4) 5400 रु.

Answer»

320-200=120120×40=4800 answer

4800 is the right answer

4800 the answer is right

Discussion

39.	0 = 16 + 4 ( m - 6 )
Answer» 0=16+4m-240=-8+4m-4m=-8m=2 16 + 4m - 24 = 0 4m = 8 m = 2

39.

0 = 16 + 4 ( m - 6 )

Answer»

0=16+4m-240=-8+4m-4m=-8m=2

16 + 4m - 24 = 0

4m = 8

m = 2

Discussion

40.	(x+a)(x+b)=x^{2}+(a+b) x+a b
Answer»

Discussion

41.	In the given figure, O is the centre of thecircle, PT is the tangent drawn from thepoint P to the circle and PAB passesthrough the centre O of the circle.If PT = 6 cm and PA = 3 cm, then find theradius of the circle.
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Discussion

42.	4) 8 -B = UIf a and b are unequal and x² + ax + b and x+ bx + a have a common factor, then
Answer» If P(x) = x^2+ ax + b and Q(x) = x^2+ bx + a have a common factor say (x - β), then P(β) = 0 and Q(β) = 0 ⇒ β^2+ aβ + b = 0 ... (1) and β^2+ bβ + a = 0 ... (2) On subtracting (1) from (2), we get (a-b)β + (b-a) = 0 ⇒ (a-b)β - (a-b) = 0 ⇒ (a-b)β = (a-b) ⇒ β = 1 On putting β = 1 in (1), we get 12+ a.1 + b = 0 ⇒ a + b = -1

42.

4) 8 -B = UIf a and b are unequal and x² + ax + b and x+ bx + a have a common factor, then

Answer»

If P(x) = x^2+ ax + b and Q(x) = x^2+ bx + a have a common factor say (x - β), then

P(β) = 0 and Q(β) = 0

⇒ β^2+ aβ + b = 0 ... (1)

and β^2+ bβ + a = 0 ... (2)

On subtracting (1) from (2), we get

(a-b)β + (b-a) = 0

⇒ (a-b)β - (a-b) = 0

⇒ (a-b)β = (a-b)

⇒ β = 1

On putting β = 1 in (1), we get

12+ a.1 + b = 0

⇒ a + b = -1

Discussion

43.	Baseitsof anareaisosceles triangle is 30 om "find
Answer» Now, Let the side = x, => x²+x² = 30²=>2x² = 30² =>x² = 1530=> x² = 53352=> x² = 5²3²2=> x = 53√2 = 15√2, now area = 1/215√215√2=> 1/2 225 * 2 => 225 sq cm, Therefore the answer is , 225cm²

43.

Baseitsof anareaisosceles triangle is 30 om "find

Answer»

Now,

Let the side = x,

=> x²+x² = 30²=>2x² = 30² =>x² = 15*30=> x² = 5*3*3*5*2=> x² = 5²*3²*2=> x = 5*3*√2 = 15√2,

now area = 1/2*15√2*15√2=> 1/2 * 225 * 2

=> 225 sq cm,

Therefore the answer is ,

225cm²

Discussion

44.	6In the fig. STOR. If PS-2.4cm, SR = 3.6cm and PQ-5 cm. Find PTPA P
Answer» Given: ST \|\| RQ PS= 3 cm SR = 4cm We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides. ar(∆PST) /ar(∆PRQ)= (PS)²/(PR)² ar(∆PST) /ar(∆PRQ)= 3²/(PS+SR)² ar(∆PST) /ar(∆PRQ)= 9/(3+4)²= 9/7²=9/49 Hence, the required ratio ar(∆PST) :ar(∆PRQ)= 9:49

44.

6In the fig. STOR. If PS-2.4cm, SR = 3.6cm and PQ-5 cm. Find PTPA P

Answer»

Given:

ST || RQ

PS= 3 cm

SR = 4cm

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

ar(∆PST) /ar(∆PRQ)= (PS)²/(PR)²

ar(∆PST) /ar(∆PRQ)= 3²/(PS+SR)²

ar(∆PST) /ar(∆PRQ)= 9/(3+4)²= 9/7²=9/49

Hence, the required ratio ar(∆PST) :ar(∆PRQ)= 9:49

Discussion

45.	13. In a parallelogram PQRS(Fig. 16.17), PQ 20 cm,QR16 cm, PT LSR,PU.LQR and PT 5.6 cm,find the length of PU.Fig. 16.17
Answer»

Discussion

46.	Using Eudid's division algorithm, find the HCF of6) 405 and 2520 u) 504 and 1188 (iüi) 960 and 1575.
Answer»

Discussion

47.	5 x+\frac{1}{x}=6 \text { let's show } 25 x^{2}+\frac{1}{x^{2}}=26
Answer» 5x+1/x=6by taking square25x^2+10+1/x^2=36so 25x^2+1/x^2=36-10=26

Discussion

48.	find the value of a and b that x^4+x^3+8x^2+ax+b is divisible by x^2+1
Answer»

Discussion

49.	tom a pofit A on the ound 960olvnm o
Answer» Like my answer if you find it useful!

Discussion

50.	10. The price of an article increases from Rs.960 to Rs.1200. Find thepercentage increase in the price
Answer» Initial price = Rs 960 After increase = Rs 1200 Increased by = 1200 - 960 = 240 % of increase = (240/960) × 100 = 25% thanks

50.

10. The price of an article increases from Rs.960 to Rs.1200. Find thepercentage increase in the price

Answer»

Initial price = Rs 960

After increase = Rs 1200

Increased by = 1200 - 960 = 240

% of increase = (240/960) × 100 = 25%

thanks

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

วิtotal surface area of a sot the to 221lid hemisphere is 462 cm2, find its volume.

CBSEsurfacearea2012of a solid hemisphere is 462 cm2, find itsCESE 2014]volume.

3. If the total surface area of a solid hemisphere is 462 cm2, find itsCBSE 2014volume.

3. If the total surface area of a solid hemisphere is 462 cm2, find its[CBSE 2014]volume.

(ii) P=き2,650,R=8% pa.andT=2years.

2The volume of a hemisphere is 24251 cm.Find its curved surface area.ICBSE 20121t a solid hemisphere is 462 cm2, find its

solid cylinder has total surface area of 462 cm2. Its curved surface area is one-third of its totalsuface area. Find the radius and height of the cylinder.A

If p12 and pa 27,find the value of p'+

If its voldme Ts T015. The total surface area of a cylinder is 462 cm2. Its curved surfaceone third of its total surface area. Find the volume of the cylinder.

20,The length12 cm andtrangle?of Mace medians of trangle are som15 cm. In exea (in sq. can of this[ons. n)

to find the radius of the circle in the trangle

then tind AOBan equilateral trangle ABC, AD is drawnProve that AD2 3BD

if trangle ABC,if AD is a meadian then show that ABsquare+ACsquare=2(ADsquare+BDsquare

12. In the figure (1 148), an equilateral trangle EAB surmounts the square ABCD Fiund the value odand yFig. 11.48

15. In the adjoining figure, ABCD is a square. A linesegment Cx cuts AB at X and the diagonal BD atO such that &lt;COD = 80° and &lt;OXA-value of xxo. Find the

What is the area of the region enclosed byy=2|xl and y = 42(a) 2 sq units (b) 4 sq units(C)8squnts (d) 16 sq units

5. In the adjoining figure, ABCD is a square. A linesegment CX cuts AB at X and the diagonal BD atO such that &lt;COD = 800 and &lt;OXA = χ。. Find thevalue of x.

8. In figure, ABCD is a rectangle. Find the value of x and y14 cm30 cm

7. It is given that in a group of 3 students, the probability of 2 students not havigy is 0.992. What is the probability that the 2 students have thesamebirthdav?

3. In ABC, if AB = 6-3 cm, AC = 12 cm and BC = 6 cm, then find the measure ofB.

1. 28 x 9 = 291 ( )2. 40 x 8 / 2 = 160 ( ) 3. 55+ 4 x 20 = 690 ( )4. 600 + 100 50 = 10650 ( )

Use Euclid's algorithm to find HCF of 1651 and 2032. Express theHCF in the form 1651m + 2032n.

If A is an event of a random experiment such thatPA): P(A)7:12,then find P(A).

PAB is a secant to the circle from a point P outside the circle. PAB passes through thecentre of the circle and PT is a tangent. If PT- 8 cm and OP 10 cm, find the radiusof the circle.

Baseitsof anareaisosceles triangle is 30 cm find

unll perimeter is 540cm. Find its area.An isosceles triangle has perimeter 30 cm and each of the equal sides is 12cm. Findthe area of the triangle..

Find the area of an isosceles triangle.

7. It is given that in a group of 3 students, the probability of 2 students not having thesame birthday is 0.992. What is the probability that the 2 students have the samebirthday?

What is the formula for area of isosceles triangle

using Euclid's algorithm to find HCF of the 1651 &amp; 2032 .express the hcf in the form 1651m+2032n

Use Euclid's algorithm to find HCF of 1190 and 1445. Express the HCFin the form 1190m - 1445n.

10. Use Euclid's algorithm to find HCF of1190and1445.ExpresstheHCFin the form 1190m + 1445n.

DatePageound the dusanctheand ma h inn a the

Find the LCM of the following numbers by short division method.a. 12, 24, 120b. 60, 72, 96c. 25, 45,95d. 27,81

48, 24, 72, 36, 108, (?)(b) 216(d) 54(c) 121

6. S27 รof 16-4-[35 รท (17 + 3 รท 4)]mpliry:3

0 = 16 + 4 ( m - 6 )

(x+a)(x+b)=x^{2}+(a+b) x+a b

In the given figure, O is the centre of thecircle, PT is the tangent drawn from thepoint P to the circle and PAB passesthrough the centre O of the circle.If PT = 6 cm and PA = 3 cm, then find theradius of the circle.

4) 8 -B = UIf a and b are unequal and x² + ax + b and x+ bx + a have a common factor, then

Baseitsof anareaisosceles triangle is 30 om "find

6In the fig. STOR. If PS-2.4cm, SR = 3.6cm and PQ-5 cm. Find PTPA P

13. In a parallelogram PQRS(Fig. 16.17), PQ 20 cm,QR16 cm, PT LSR,PU.LQR and PT 5.6 cm,find the length of PU.Fig. 16.17

Using Eudid's division algorithm, find the HCF of6) 405 and 2520 u) 504 and 1188 (iüi) 960 and 1575.

5 x+\frac{1}{x}=6 \text { let's show } 25 x^{2}+\frac{1}{x^{2}}=26

find the value of a and b that x^4+x^3+8x^2+ax+b is divisible by x^2+1

tom a pofit A on the ound 960olvnm o

10. The price of an article increases from Rs.960 to Rs.1200. Find thepercentage increase in the price

15. In the adjoining figure, ABCD is a square. A linesegment Cx cuts AB at X and the diagonal BD atO such that <COD = 80° and <OXA-value of xxo. Find the

5. In the adjoining figure, ABCD is a square. A linesegment CX cuts AB at X and the diagonal BD atO such that <COD = 800 and <OXA = χ。. Find thevalue of x.

using Euclid's algorithm to find HCF of the 1651 & 2032 .express the hcf in the form 1651m+2032n

6. S27 รof 16-4-[35 รท (17 + 3 รท 4)]mpliry:3