81279 + Interview Questions in Mathematics QA in Mathematics

1.	6. The volume of a right circular cone is 9856 cm. If the diameter of the base is 28 cm,find(i) height of the cone(ii) curved surface area of the cone(ii) slant height of the coneider 5cm 12 cmand 13 cm is revolved about the side 12 cm.
Answer» thanks

Discussion

2.	VIL IF25% of a number is 1200, then find the number.
Answer» right answer is 4800 The right answer is 4800 4800 is right answer 4800 is correct answer the correct answer 4800 4800 is a right answer the correct answer is 4800 25percent of x =120025x x =1200100x=1200x100 5=4800 300 is the answer... 4800.... is the answer 4800 is the best answer me x=4800 is the best answer X=4800 will be your answer friend 25% of X=120025/100 x X=120025X/100=120025X =120000X=120000/25X=4800 Ans. 4800 is the right answer 4800 is correct answet 25% of x= 120025/100 ×x=.1200X= 1200×100 25=48×100=4800 x=4800 is the correct answer 25% of x = 120025/100 × x = 1200x = 4800 25 percent of a number is 1200 lets consider the number as x25 percent = 25/10025/100 * x = 120025/100x = 120025x = 1200*10025x = 120000x = 120000/5x = 4800

2.

VIL IF25% of a number is 1200, then find the number.

Answer»

right answer is 4800

The right answer is 4800

4800 is right answer

4800 is correct answer

the correct answer 4800

4800 is a right answer

the correct answer is 4800

25percent of x =120025x x =1200100x=1200x100 5=4800

300 is the answer...

4800.... is the answer

4800 is the best answer me

x=4800 is the best answer

X=4800 will be your answer friend

25% of X=120025/100 x X=120025X/100=120025X =120000X=120000/25X=4800 Ans.

4800 is the right answer

4800 is correct answet

25% of x= 120025/100 ×x=.1200X= 1200×100 25=48×100=4800

x=4800 is the correct answer

25% of x = 120025/100 × x = 1200x = 4800

25 percent of a number is 1200

lets consider the number as x25 percent = 25/10025/100 * x = 120025/100x = 120025x = 1200*10025x = 120000x = 120000/5x = 4800

Discussion

3.	Find area of the triangle if its base is 7cm and perpendicular on the base is 5cm.
Answer»

Discussion

4.	Find area of the triangle if its base is 7cm and perpendicular on the base is 5cm.
Answer»

Discussion

5.	BAT12 UOLY ~ SO Vil 226ÂĽ _ SO .y g +s & Dbl
Answer» SOLUTION: In ΔPQR, ∠1 = ∠2∠PQR = ∠PRQ [GIVEN]∴ PR = PQ ……………..…(1) [Sides opposite to equal angles of a triangle are also equal]Given: QR/QS = QT/PRQR/QS = QT/PQ QS/QR = PQ/QT…... …(ii) [Taking reciprocals] [From eq (i)]In ΔPQS and ΔTQR,QS/QR = PQ/QT [From eq (ii)] ∠PQS = ∠TQR [ common]∴ ΔPQS ~ ΔTQR [By SAS similarity criterion] Hence, proved hit like if you find it useful

5.

BAT12 UOLY ~ SO Vil 226ÂĽ _ SO .y g +s & Dbl

Answer»

SOLUTION:

In ΔPQR, ∠1 = ∠2∠PQR = ∠PRQ [GIVEN]∴ PR = PQ ……………..…(1)

[Sides opposite to equal angles of a triangle are also equal]Given: QR/QS = QT/PRQR/QS = QT/PQ

QS/QR = PQ/QT…... …(ii) [Taking reciprocals]

[From eq (i)]In ΔPQS and ΔTQR,QS/QR = PQ/QT [From eq (ii)] ∠PQS = ∠TQR [ common]∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]

Hence, proved

hit like if you find it useful

Discussion

6.	5. Find area of the triangle if its base is 7cm and perpendicular on the base is 5cm.
Answer» Area of traingle = 1/2 × base × height = 1/2 × 7 × 5 cm² = 35/2 = 17.5 cm²

Discussion

7.	.Find the supplement of each of the given angles. 45째2. 95째3. 160째4. 125째W/uito T far True or F for False.
Answer» Supplement=180 1. 180-45=1352. 180-95=853. 180-160=204. 180-125=55

7.

.Find the supplement of each of the given angles. 45째2. 95째3. 160째4. 125째W/uito T far True or F for False.

Answer»

Supplement=180

1. 180-45=1352. 180-95=853. 180-160=204. 180-125=55

Discussion

8.	In the figure 10.9, l 11 m II n and p is a transversal. If L1 = 110°, find angles x, y, z,and w.1.110°> nFig. 10.9
Answer» x = 180° -110° = 70°y = 70° ( vertically opposite angle)z = 70° (corresponding angle)w = 180-70 = 110° ( linear pair)

Discussion

9.	1 Identify the next term in the follownumber series220, 284, 356,?(A) 400(B) 411(C) 436(D) 441
Answer» 436 option c is the right answer option c is right answer 436

9.

1 Identify the next term in the follownumber series220, 284, 356,?(A) 400(B) 411(C) 436(D) 441

Answer»

436 option c is the right answer

option c is right answer 436

Discussion

10.	find the surface area of the cuboid measuring 3m×2m×1.5m
Answer» no answer kiddingkiddingOregonkusum

Discussion

11.	find the first five term of following given sequences: am(2m) - 3m +
Answer» x^2-5x+4=0; x^2-4x-x+4=0;; x( x-4)-1( x-4)=0; ( x-4)( x-1); x=4,1

Discussion

12.	2m/7+3m/4=17
Answer» 2m/7 + 3m/4 = 17Take LCM (8m + 21m)/28 = 1729m = 17*28m = 16.41

12.

2m/7+3m/4=17

Answer»

2m/7 + 3m/4 = 17Take LCM

(8m + 21m)/28 = 1729m = 17*28m = 16.41

Discussion

13.	1. Find the area of a \|m with(i)(ii)base 9cm and height 5cmbase 2m and height 40cm.base 3m 4dm and height 2m 3dm
Answer»

Discussion

14.	3. Two ropes are 51/5 m and 61/5m long. Find the3total length of rope.
Answer» Thankyou apne mere question solve kar diya

Discussion

15.	3.Divide the following expressions, and write2m(m2 12m +35)+ 3m(m 7)(b) 12at
Answer»

Discussion

16.	\int e^{x} d x=e^{x}
Answer»

Discussion

17.	dtionation \| Cue t6om ciimuy o Sioma (keLaneacocte Ak Mpcod by 2Se Yonlta dne itsnual Apcecl Blavd 436 usual Speasds\| Ay NI
Answer» Given: Late = 30 min Distance = 1500 km Speed increase = 100 km / hr To find: The usual speed. Solution: By formula, Speed = Distance / Time Since the time is unknown, Consider x for time, Speed original = 1500 / x Time taken = ( x - 1 / 2 ) Therefore, Substituting the value of x, Speed new = 1500 / (x - 1/2) Speed New - Speed original = 100 ( 1500 / ( x - 1 / 2 ) ) - ( 1500 / x ) = 100 150 ( ( 10 / x - 1 / 2 ) - 10 / x = 100 30 x -30 x +15 = 2x^2 - x 15 = 2x^2 - x 2x^2 - x -15 = 0 Solving, x = -5/2 x = 3 As time cannot be negative, The usual time taken by the plane is 3 hr Speed original = ( 1500 / 3 ) = 500 km / hr Hence, the usual speed of the plane is 500 km / hr

17.

dtionation | Cue t6om ciimuy o Sioma (keLaneacocte Ak Mpcod by 2Se Yonlta dne itsnual Apcecl Blavd 436 usual Speasds| Ay NI

Answer»

Given:

Late = 30 min

Distance = 1500 km

Speed increase = 100 km / hr

To find:

The usual speed.

Solution:

By formula,

Speed = Distance / Time

Since the time is unknown,

Consider x for time,

Speed original = 1500 / x

Time taken = ( x - 1 / 2 )

Therefore,

Substituting the value of x,

Speed new = 1500 / (x - 1/2)

Speed New - Speed original = 100

( 1500 / ( x - 1 / 2 ) ) - ( 1500 / x ) = 100

150 ( ( 10 / x - 1 / 2 ) - 10 / x = 100

30 x -30 x +15 = 2x^2 - x

15 = 2x^2 - x

2x^2 - x -15 = 0

Solving,

x = -5/2

x = 3

As time cannot be negative,

The usual time taken by the plane is 3 hr

Speed original = ( 1500 / 3 ) = 500 km / hr

Hence, the usual speed of the plane is 500 km / hr

Discussion

18.	Dne angle of a triangle is oFind the romaining w angles if their diffieence
Answer» 180 = 65 + x + x + 20 2x = 95 x = 47.5 The angles are 47.5 and 67.5

18.

Dne angle of a triangle is oFind the romaining w angles if their diffieence

Answer»

180 = 65 + x + x + 20

2x = 95

x = 47.5

The angles are 47.5 and 67.5

Discussion

19.	\int e ^ { - \operatorname { log } _ { e } x } d x
Answer» v good

Discussion

20.	\int ( e ^ { x } + 1 ) ^ { 2 } e ^ { x } d x
Answer»

Discussion

21.	\int e ^ { x } \cdot \operatorname { cos } ( e ^ { x } ) d x
Answer» ∫ e^x cos(e^x) dx let e^x = t e^x dx = dtso∫ e^x cos(e^x) dx will be∫ cos(t) dt = sin(t) + c so answer is sin(e^x) +c

21.

\int e ^ { x } \cdot \operatorname { cos } ( e ^ { x } ) d x

Answer»

∫ e^x cos(e^x) dx

let e^x = t e^x dx = dtso∫ e^x cos(e^x) dx will be∫ cos(t) dt = sin(t) + c

so answer is sin(e^x) +c

Discussion

22.	poster on mathamatics in architecture
Answer»

Discussion

23.	\int e^{x} \cdot \cos \left(e^{x}\right) d x
Answer»

Discussion

24.	यदि am 5 2m* + 3m + 6, 3m? 4 4 4 Uh U o मांगते पद+8 < ) ७ m एक स पंतर श्रेढ़ी के तीन क्र 4= लो #का मान ज्ञात कीरि जिए \| O
Answer»

Discussion

25.	( x i v ) \int ( e ^ { a \operatorname { log } x } + e ^ { x \operatorname { log } a } ) d x
Answer»

Discussion

26.	3. The volume of a cube is 343 cm 3. Its total surface area is
Answer»

Discussion

27.	3. The volume of a cube is 343 cm3. Its total surface area is
Answer»

Discussion

28.	8. Three pieces of ropes measuring 0.2 km, 5.005 cm and 1,200 mm respectively are joinedend to end to form a single rope.(a) How long is the new rope?(b) Find the cost of the single rope so formed at the rate of Rs. 15 per metre.
Answer» length of new rope is 25125 cm.cost is 3768.75₹

Discussion

29.	56PQRS is a kite and R lies on the line XY. Iffollowing:(a) LQRS(b) QPSQRX-250 and LSRY-35, find the measures ofe(c)If<S=60°, find ZQ.2535。
Answer» angle qrs = 180-25-35 = 120°

29.

56PQRS is a kite and R lies on the line XY. Iffollowing:(a) LQRS(b) QPSQRX-250 and LSRY-35, find the measures ofe(c)If<S=60°, find ZQ.2535。

Answer»

angle qrs = 180-25-35

= 120°

Discussion

30.	8. Three pieces of ropes measuring 0.2 km, 5,005 cm and 1,200 mm respectively are joinedend to end to form a single rope.(a) How long is the new rope?(b) Find the cost of the single rope so formed at the rate of Rs. 15 per metre.
Answer» 1km. Equal to 5 rupees. And 5 km equal to25 rupees per month rupee=0.2km=20000 cm, 5,005= 5005 CM 1800mm=120 cm, new rupee 20,000+5005+120 cm, 1)the new rupee ii)25.125 cm long 2)rs.15 per meter=251.25 cm rupee 251.25 meter ×15= 3.768.75

30.

8. Three pieces of ropes measuring 0.2 km, 5,005 cm and 1,200 mm respectively are joinedend to end to form a single rope.(a) How long is the new rope?(b) Find the cost of the single rope so formed at the rate of Rs. 15 per metre.

Answer»

1km. Equal to 5 rupees. And 5 km equal to25 rupees per month

rupee=0.2km=20000 cm, 5,005= 5005 CM 1800mm=120 cm, new rupee 20,000+5005+120 cm, 1)the new rupee ii)25.125 cm long 2)rs.15 per meter=251.25 cm rupee 251.25 meter ×15= 3.768.75

Discussion

31.	we y = Secontannetdne(A-Single
Answer»

Discussion

32.	164 in the given figure, Show that XY!\|EF.
Answer»

Discussion

33.	16. List 2 integers smaller than-3 but their difference is greater than -3.
Answer» The two integers can be -4 and -5 as, (-4-(-5))=1 and (-5-(-4))=-1 are both greater than -3.

Discussion

34.	16. List 2 integers smaller than -3 but their difference is greater than-3.
Answer» The two integers can be -4 and -5 as, (-4-(-5))=1 and (-5-(-4))=-1 are both greater than -3.

Discussion

35.	0,1,2,3 ko use krne char anko ki sbse chhoti snkhya bnaiye
Answer» 1023 is the smallest 4 digit nmbr 1023 is the correct answer of the given question the right answer is 0123 1023 is right answer of this question Smallest number by using digits 0,1,2,3 is 1023. 1023 is the smallest 4digit no.that can be made

35.

0,1,2,3 ko use krne char anko ki sbse chhoti snkhya bnaiye

Answer»

1023 is the smallest 4 digit nmbr

1023 is the correct answer of the given question

the right answer is 0123

1023 is right answer of this question

Smallest number by using digits 0,1,2,3 is 1023.

1023 is the smallest 4digit no.that can be made

Discussion

36.	\left. \begin{array} { l } { I = \int \sqrt { 1 + \operatorname { sin } x } d x , \text { then } } \\ { - 2 \sqrt { 1 - \operatorname { sin } x } + C } \\ { \operatorname { sin } ( x / 2 ) + \operatorname { cos } ( x / 2 ) + C } \\ { \operatorname { cos } ( x / 2 ) - \operatorname { cos } ( x / 2 ) + C } \\ { 2 \sqrt { 1 - \operatorname { sin } x } + C } \end{array} \right.
Answer» Could you please find its integration.

Discussion

37.	I = \int \frac { e ^ { x } ( 1 + \operatorname { sin } x ) } { ( 1 + \operatorname { cos } x ) } d x = \int e ^ { x } \frac { 1 + 2 \operatorname { sin } \frac { x } { 2 } \cdot \operatorname { cos } \frac { x } { 2 } } { 2 \operatorname { cos } ^ { 2 } \frac { x } { 2 } } d x
Answer»

Discussion

38.	10xin terms of a and b.80) If log x ะเ + b and logy-a-b, express log lyīj
Answer» thank you very much

Discussion

39.	4.Express log 48 in terms of log 2 and log 3.
Answer» Log (48) = Log(6×8)= log(2^4×3)so 4log2+log3

Discussion

40.	Use Euclid's division lemma to show that the square of any positive integer is either ofthe form 3m or 3m + 1 for some integer m
Answer»

Discussion

41.	Express $ 2 \log 3-\frac{1}{2} \log 16+\log 12 $ as a single Logaruthen
Answer»

Discussion

42.	16. In the given figure , show that XY// EF5630
Answer» angleXYR =36 +60 =66 so, in thw given figure show that xy!!ef

42.

16. In the given figure , show that XY// EF5630

Answer»

angleXYR =36 +60 =66 so,

in thw given figure show that xy!!ef

Discussion

43.	Find x in the given figures.x = inches.4816çł
Answer»

Discussion

44.	find the GCF for the given list 16, 40The GCF is
Answer» To get the Greates Common Factor (GCF) of 16 and 40 we need to factor each value first and then we choose all the copies of factors and multiply them: 16: 222240: 2225GCF: 222The Greates Common Factor (GCF) is: 2 x 2 x 2 = 8

44.

find the GCF for the given list 16, 40The GCF is

Answer»

To get the Greates Common Factor (GCF) of 16 and 40 we need to factor each value first and then we choose all the copies of factors and multiply them:

16: 222240: 2225GCF: 222The Greates Common Factor (GCF) is: 2 x 2 x 2 = 8

Discussion

45.	HCF of 144 180 and 192
Answer» 144=2^43^2180=2^23^25192=263so HCF=2^23=4*3=12 I didn't get you Thank you for your try

45.

HCF of 144 180 and 192

Answer»

144=2^4*3^2180=2^2*3^2*5192=2*6*3so HCF=2^2*3=4*3=12

I didn't get you

Thank you for your try

Discussion

46.	IV. In the space given below, draw the pictures ofany two objects that you find in your house.
Answer» I am t to get a new phone and the phone number is the same as the one you sent to the address you have on file the claim and it

Discussion

47.	Thus, tne aiode is 10,000.ILLUSTRATION14. (b) The difference between CI. and S.I. is 999 when the ratef investment.is 10% p.a. for a period of 5 years, then find the amount oGiven (1.1ćś-1.611.
Answer» SI = PRT/100 = P105/100 = P/2 For Compound interesrtA = P(1 + R/100)^5 = P(1 + 10/100)^5 = P(1.1)^5 = 1.611P CI = A - P = 1.611P - P =.611P Given CI - SI = 999 .611P -.5P = 999 .111P = 999 P = 900 Amount of investment = Rs 900

47.

Thus, tne aiode is 10,000.ILLUSTRATION14. (b) The difference between CI. and S.I. is 999 when the ratef investment.is 10% p.a. for a period of 5 years, then find the amount oGiven (1.1ćś-1.611.

Answer»

SI = P*R*T/100 = P*10*5/100 = P/2

For Compound interesrtA = P(1 + R/100)^5 = P(1 + 10/100)^5 = P(1.1)^5 = 1.611P

CI = A - P = 1.611P - P =.611P

Given CI - SI = 999 .611P -.5P = 999 .111P = 999 P = 900

Amount of investment = Rs 900

Discussion

48.	6. The average age of 20 boys is 14 yr. A boy of 16 leaves them and a new boy occupieshis place. Now their average is 13.8 yr. What is the age of the newcomer?
Answer»

Discussion

49.	, FactorieThe area of a...parallëlograim is 240 cm2 andaparalelogan is 240 cme anddistance betveen parallel side is 16cm. Find thebase.
Answer»

Discussion

50.	if HCF of 144 and 180 is express in the form 3m-3 find the value of m
Answer»

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

6. The volume of a right circular cone is 9856 cm. If the diameter of the base is 28 cm,find(i) height of the cone(ii) curved surface area of the cone(ii) slant height of the coneider 5cm 12 cmand 13 cm is revolved about the side 12 cm.

VIL IF25% of a number is 1200, then find the number.

Find area of the triangle if its base is 7cm and perpendicular on the base is 5cm.

Find area of the triangle if its base is 7cm and perpendicular on the base is 5cm.

BAT12 UOLY ~ SO Vil 226ÂĽ _ SO .y g +s &amp; Dbl

5. Find area of the triangle if its base is 7cm and perpendicular on the base is 5cm.

.Find the supplement of each of the given angles. 45째2. 95째3. 160째4. 125째W/uito T far True or F for False.

In the figure 10.9, l 11 m II n and p is a transversal. If L1 = 110°, find angles x, y, z,and w.1.110°&gt; nFig. 10.9

1 Identify the next term in the follownumber series220, 284, 356,?(A) 400(B) 411(C) 436(D) 441

find the surface area of the cuboid measuring 3m×2m×1.5m

find the first five term of following given sequences: am(2m) - 3m +

2m/7+3m/4=17

1. Find the area of a |m with(i)(ii)base 9cm and height 5cmbase 2m and height 40cm.base 3m 4dm and height 2m 3dm

3. Two ropes are 51/5 m and 61/5m long. Find the3total length of rope.

3.Divide the following expressions, and write2m(m2 12m +35)+ 3m(m 7)(b) 12at

\int e^{x} d x=e^{x}

dtionation | Cue t6om ciimuy o Sioma (keLaneacocte Ak Mpcod by 2Se Yonlta dne itsnual Apcecl Blavd 436 usual Speasds| Ay NI

Dne angle of a triangle is oFind the romaining w angles if their diffieence

\int e ^ { - \operatorname { log } _ { e } x } d x

\int ( e ^ { x } + 1 ) ^ { 2 } e ^ { x } d x

\int e ^ { x } \cdot \operatorname { cos } ( e ^ { x } ) d x

poster on mathamatics in architecture

\int e^{x} \cdot \cos \left(e^{x}\right) d x

यदि am 5 2m* + 3m + 6, 3m? 4 4 4 Uh U o मांगते पद+8 &lt; ) ७ m एक स पंतर श्रेढ़ी के तीन क्र 4= लो #का मान ज्ञात कीरि जिए | O

( x i v ) \int ( e ^ { a \operatorname { log } x } + e ^ { x \operatorname { log } a } ) d x

3. The volume of a cube is 343 cm 3. Its total surface area is

3. The volume of a cube is 343 cm3. Its total surface area is

8. Three pieces of ropes measuring 0.2 km, 5.005 cm and 1,200 mm respectively are joinedend to end to form a single rope.(a) How long is the new rope?(b) Find the cost of the single rope so formed at the rate of Rs. 15 per metre.

56PQRS is a kite and R lies on the line XY. Iffollowing:(a) LQRS(b) QPSQRX-250 and LSRY-35, find the measures ofe(c)If&lt;S=60°, find ZQ.2535。

8. Three pieces of ropes measuring 0.2 km, 5,005 cm and 1,200 mm respectively are joinedend to end to form a single rope.(a) How long is the new rope?(b) Find the cost of the single rope so formed at the rate of Rs. 15 per metre.

we y = Secontannetdne(A-Single

164 in the given figure, Show that XY!|EF.

16. List 2 integers smaller than-3 but their difference is greater than -3.

16. List 2 integers smaller than -3 but their difference is greater than-3.

0,1,2,3 ko use krne char anko ki sbse chhoti snkhya bnaiye

I = \int \frac { e ^ { x } ( 1 + \operatorname { sin } x ) } { ( 1 + \operatorname { cos } x ) } d x = \int e ^ { x } \frac { 1 + 2 \operatorname { sin } \frac { x } { 2 } \cdot \operatorname { cos } \frac { x } { 2 } } { 2 \operatorname { cos } ^ { 2 } \frac { x } { 2 } } d x

10xin terms of a and b.80) If log x ะเ + b and logy-a-b, express log lyīj

4.Express log 48 in terms of log 2 and log 3.

Use Euclid's division lemma to show that the square of any positive integer is either ofthe form 3m or 3m + 1 for some integer m

Express $ 2 \log 3-\frac{1}{2} \log 16+\log 12 $ as a single Logaruthen

16. In the given figure , show that XY// EF5630

Find x in the given figures.x = inches.4816çł

find the GCF for the given list 16, 40The GCF is

HCF of 144 180 and 192

IV. In the space given below, draw the pictures ofany two objects that you find in your house.

Thus, tne aiode is 10,000.ILLUSTRATION14. (b) The difference between CI. and S.I. is 999 when the ratef investment.is 10% p.a. for a period of 5 years, then find the amount oGiven (1.1ćś-1.611.

6. The average age of 20 boys is 14 yr. A boy of 16 leaves them and a new boy occupieshis place. Now their average is 13.8 yr. What is the age of the newcomer?

, FactorieThe area of a...parallëlograim is 240 cm2 andaparalelogan is 240 cme anddistance betveen parallel side is 16cm. Find thebase.

if HCF of 144 and 180 is express in the form 3m-3 find the value of m

BAT12 UOLY ~ SO Vil 226ÂĽ _ SO .y g +s & Dbl

In the figure 10.9, l 11 m II n and p is a transversal. If L1 = 110°, find angles x, y, z,and w.1.110°> nFig. 10.9

यदि am 5 2m* + 3m + 6, 3m? 4 4 4 Uh U o मांगते पद+8 < ) ७ m एक स पंतर श्रेढ़ी के तीन क्र 4= लो #का मान ज्ञात कीरि जिए | O

56PQRS is a kite and R lies on the line XY. Iffollowing:(a) LQRS(b) QPSQRX-250 and LSRY-35, find the measures ofe(c)If<S=60°, find ZQ.2535。

Find x in the given figures.x = inches.4816çł

Thus, tne aiode is 10,000.ILLUSTRATION14. (b) The difference between CI. and S.I. is 999 when the ratef investment.is 10% p.a. for a period of 5 years, then find the amount oGiven (1.1ćś-1.611.