81279 + Interview Questions in Mathematics QA in Mathematics Page 61 InterviewSolution

3001.	Two cylindrical vessels are filled with milk. The radius of one vessel is 15 cm andheight is 40 cm, and the radius of other vessel is 20 cm and height is 45 cm. Find theradius of another cylindrical vessel of height 30 em which may just contain the milkwhich is in the two given vessels.
Answer»

Discussion

3002.	areaBase18 cmHeight15cm
Answer» Area= 1/2 x base x height= 1/2* 18 * 5= 45cm ^2 area=1/2×base×hight=1/2×18×5=45cm² answer area =1/2baseheight=1/2185=45 if it is parallelogram, area = baseheight=85=90 area =1/2 * base * height = 1/2* 18* 5=. 45 cm sq area =1/2base height =1/2* 185=.45cm sq Area = 1/2 × b x h = 1/2 × 18 x 5 = 9x5=45cm^2 area=1/2×base×height =1/2×18×5 =45cm^2 area of triangle=1/2×base×height=1/2×18×5=9×5=45 cm^2. ans. area = 45 cm^2 is a correct answer area=1/2 base * height 1/2* 18* 5= 45 cm sq area = 1/2basehight=1/2185=45cm sq 45 cm2 is the answer

3002.

areaBase18 cmHeight15cm

Answer»

Area= 1/2 x base x height= 1/2* 18 * 5= 45cm ^2

area=1/2×base×hight=1/2×18×5=45cm² answer

area =1/2*base*height=1/2*18*5=45

if it is parallelogram, area = base*height=8*5=90

area =1/2 * base * height = 1/2* 18* 5=. 45 cm sq

area =1/2*base *height =1/2* 18*5=.45cm sq

Area = 1/2 × b x h = 1/2 × 18 x 5 = 9x5=45cm^2

area=1/2×base×height =1/2×18×5 =45cm^2

area of triangle=1/2×base×height=1/2×18×5=9×5=45 cm^2. ans.

area = 45 cm^2 is a correct answer

area=1/2 * base * height 1/2* 18* 5= 45 cm sq

area = 1/2*base*hight=1/2*18*5=45cm sq

45 cm2 is the answer

Discussion

3003.	.उदाहरण 15. सिद्ध कीजिए कि:ca 30.
Answer» multiply 1st column with abc you will get the variables as in 3rd column now subtract c1-c3 then c1 = 0 any column is zero then total det is equals to 0

Discussion

3004.	A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. Thisucket is emptied on the ground and a conical heap of sand is formed. If the height of theconical heap is 24 cm, find the radius and slant height of the heap.
Answer»

Discussion

3005.	A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand Tbucket is emptied on the ground and a conical heap of sand is formedconical heap is 24 cm, find the radius and slant height of the heap.If the height of's
Answer»

Discussion

3006.	7. Acylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. Thisbucket is emptied on the ground and a conical heap of sand is formed. Ifthe height of theconical heap is 24 cm, find the radius and slant height of the heap.
Answer»

Discussion

3007.	A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. Thisbucket is emptied on the ground and a conical heap of sand is formed. If the height of theconical heap is 24 cm, find the radius and slant height of the heap.
Answer» thanks

Discussion

3008.	7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled withand. Thibucket is emptied on the ground and a conical heap of sand is formed. If the height of theconical heap is 24 cm, find the radius and slant height of the heap.
Answer»

Discussion

3009.	20. Determine the missing frequency x, from thefollowing data, when mode is 67.Class40-50 50-60 -60-70 70-80 80-90Frequency5x 15127
Answer»

Discussion

3010.	7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filledThnd is formed. If the height of thebucket is emptied on the ground and a conical heap of sand is formed. Ifthconical heap is 24 cm, find the radius and slant height of the heap
Answer»

Discussion

3011.	The Class marks of a distribution are 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97, and 102.Determine the Class size?
Answer»

Discussion

3012.	5.The Class marks of a distribution are 47,52, 57, 62, 67, 72, 77, 82, 87, 92, 97, and 102.Determine the Class size?
Answer»

Discussion

3013.	x2+ 18x +81
Answer»

Discussion

3014.	(7t %2B 2)/(-5t %2B 1)=(-7t %2B 3)/(5t %2B 4)
Answer» nahi aaraha ha bhai next question bolo

Discussion

3015.	(a)18x(7 + (-3)] = [18×7] + [ 18 × (-3)]
Answer» Yes it is equal due to distributive property of multiplication

Discussion

3016.	e oooके=0 z ¢v o Yo c"* !«}3"45 .QY पी हिa8 Qe: QN _‘j Nहि १(दी १\|)
Answer» Properties of a parallelogram. 1)Diagonalsbisecteach other and each diagonal divides the parallelogram into twocongruenttriangles.2) If one of the angles of a parallelogram is a right angle then all other angles are right and it becomes a rectangle

Discussion

3017.	4x^4+x^3-73x^2-18x divide by x^2-18
Answer»

Discussion

3018.	) 18 рек [7 + (-3)] * [18 * 7] + [18 x (-3)]..... 4 MBS &
Answer» 18[ 7 + (-3)] = [18×7] + [18×(-3)] = 126 - 54 = 72 If you find this answer helpful then like it.

3018.

) 18 рек [7 + (-3)] * [18 * 7] + [18 x (-3)]..... 4 MBS &

Answer»

18[ 7 + (-3)] = [18×7] + [18×(-3)] = 126 - 54 = 72

If you find this answer helpful then like it.

Discussion

3019.	18x+18x=72 then x=?
Answer» x=2 is the right answer

Discussion

3020.	Obtain other zeroes of the polynomial 4xx-x-18x, if two of its zeroes are 32 and-3/2
Answer»

Discussion

3021.	18x=13x+62
Answer» 18x-13x = 62 5x = 62 x = 62/5 thanks

3021.

18x=13x+62

Answer»

18x-13x = 62

5x = 62

x = 62/5

thanks

Discussion

3022.	(21(text(18(a(dn)))))(-3 %2B 21) %2B 18*(-3)
Answer» -117 and 324...is write -54+21 and 18×18-33 and 324 plz accept as best both is -33 both is equal -33 and 324 is the correct answer of the given question 18×(-3)+21=(-54)+21=(-33) 18×[(-3)+21]=18×18=324Both are not equal.

3022.

(21(text(18(a(dn)))))(-3 %2B 21) %2B 18*(-3)

Answer»

-117 and 324...is write

-54+21 and 18×18-33 and 324

plz accept as best

both is -33 both is equal

-33 and 324 is the correct answer of the given question

18×(-3)+21=(-54)+21=(-33)

18×[(-3)+21]=18×18=324Both are not equal.

Discussion

3023.	7. If θ(i)30°, verify thatsin 2θ2 sin θ cos θ
Answer»

Discussion

3024.	If $ \theta=30^{\circ}, $ verify that$ 1-\cos 2 \theta=2 \sin ^{2} \theta $
Answer»

Discussion

3025.	1/13 %2B 3sqrt(3) %2B 2(7*sqrt(2))
Answer»

Discussion

3026.	40.Current I in the circuit will be :-20Ω30 Ω20 Ω5 V5(1)a A(2) 50 A(3)to AOA(4)
Answer»

Discussion

3027.	a) 65x13
Answer» 65×13 = 13×13×5 = 169×5 = 845

Discussion

3028.	\frac { 4 \sqrt { 3 } } { 2 - \sqrt { 2 } } - \frac { 30 } { 4 \sqrt { 3 } - \sqrt { 18 } } - \frac { \sqrt { 18 } } { 3 - \sqrt { 12 } }
Answer» thanks

Discussion

3029.	\frac{4 \sqrt{3}}{2-\sqrt{2}}-\frac{30}{4 \sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}
Answer» [1] = (4√3) / (2-√2) = (4√3)(2+√2) / (2-√2)(2+√2) = (8√3+4√6) /2 = 4√3+2√6 …………………(1) [2]= 30 / (4√3-√18) = 30*(4√3+√18) /(4√3-√18)(4√3+√18) =( 120√3 +30√18) /(48–18) = 30(4√3+√18) /30 = 4√3 + 3√2 ………………(2) [3]= √18 / (3–2√3) = √18(3+2√3) / (3–2√3) (3+2√3) = (3√18+2√54) /9–12 = (9√2 + 6√6) /-3 = -3√2 -2√6 ………………(3) Now by combining all 3 terms: (4√3+2√6) — (4√3+3√2) — ( -3√2 -2√6) = 4√3+2√6–4√3–3√2+3√2+2√6 = 2√6 + 2√6 = 4√6

3029.

\frac{4 \sqrt{3}}{2-\sqrt{2}}-\frac{30}{4 \sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}

Answer»

[1] = (4√3) / (2-√2) = (4√3)(2+√2) / (2-√2)(2+√2)

= (8√3+4√6) /2

= 4√3+2√6 …………………(1)

[2]= 30 / (4√3-√18)

= 30*(4√3+√18) /(4√3-√18)(4√3+√18)

=( 120√3 +30√18) /(48–18)

= 30(4√3+√18) /30

= 4√3 + 3√2 ………………(2)

[3]= √18 / (3–2√3)

= √18(3+2√3) / (3–2√3) (3+2√3)

= (3√18+2√54) /9–12

= (9√2 + 6√6) /-3

= -3√2 -2√6 ………………(3)

Now by combining all 3 terms:

(4√3+2√6) — (4√3+3√2) — ( -3√2 -2√6)

= 4√3+2√6–4√3–3√2+3√2+2√6

= 2√6 + 2√6

= 4√6

Discussion

3030.	+ bv/3, then find a and(3)If (5W2 + 3:3) _ (6V2-7:3) = a12
Answer»

Discussion

3031.	8. Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.
Answer» As angle of recatngle is 90°we will apply pythaogras theorem 35² + 12² = Digonal²1369 = digonal²37² = Digonal²Digonal = 37

Discussion

3032.	30525 aI8\frac{4 \sqrt{3}}{2-\sqrt{2}}-\frac{30}{4 \sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}
Answer»

Discussion

3033.	2. Solve;(c) 77318 18 (15 15+3 (g)3
Answer» I can't understand (G) of question 2 solve I can't understand (G) of question 2 solve

3033.

2. Solve;(c) 77318 18 (15 15+3 (g)3

Answer»

I can't understand (G) of question 2 solve

Discussion

3034.	If $ ^{18} C_{15}+2^{18} C_{16}+17 C_{16}+1=^{n} C_{3} $ then(1) 19(3) 18
Answer»

Discussion

3035.	(a) 18 x [7+ (-3)] [18 x 7]+ [18x (-3)]
Answer» 18[ 7 + (-3)] = [18×7] + [18×(-3)] = 126 - 54 = 72 If you find this answer helpful then like it.

3035.

(a) 18 x [7+ (-3)] [18 x 7]+ [18x (-3)]

Answer»

18[ 7 + (-3)] = [18×7] + [18×(-3)] = 126 - 54 = 72

If you find this answer helpful then like it.

Discussion

3036.	\frac { 5 ^ { n + 2 } - 6 \times 5 ^ { n + 1 } } { 13 \times 5 ^ { n } - 2 \times 5 ^ { n + 1 } } Q
Answer» =5ⁿ5²-65ⁿ5/135ⁿ-25ⁿ5 =5ⁿ(5²-65)/5ⁿ(13-25) =5²-65/13-25 =25-30/13-10 =-5/3 where does the 5to the power n goes in third step

3036.

\frac { 5 ^ { n + 2 } - 6 \times 5 ^ { n + 1 } } { 13 \times 5 ^ { n } - 2 \times 5 ^ { n + 1 } } Q

Answer»

=5ⁿ*5²-6*5ⁿ*5/13*5ⁿ-2*5ⁿ*5

=5ⁿ(5²-6*5)/5ⁿ(13-2*5)

=5²-6*5/13-2*5

=25-30/13-10

=-5/3

where does the 5to the power n goes in third step

Discussion

3037.	\frac { 5 ^ { n + 2 } - 6 \times 5 ^ { n + 1 } } { 13 \times 5 ^ { n } - 2 \times 5 ^ { n + 1 } }
Answer» =5n52-65n51/135n-25n51 =5n(52-651)/5n(13-251) =52-65/13-25 =25-30/13-10 =-5/3

3037.

\frac { 5 ^ { n + 2 } - 6 \times 5 ^ { n + 1 } } { 13 \times 5 ^ { n } - 2 \times 5 ^ { n + 1 } }

Answer»

=5n*52-6*5n*51/13*5n-2*5n*51

=5n(52-6*51)/5n(13-2*51)

=52-6*5/13-2*5

=25-30/13-10

=-5/3

Discussion

3038.	\frac{10 \times 5^{n+1}+25 \times 5^{n}}{3 \times 5^{n+2}+10 \times 5^{n+1}}
Answer»

Discussion

3039.	(-65^(n %2B 1) %2B 5^(n %2B 2))/(135^n - 2*5^(n %2B 1))
Answer» _ 5/3 is the right answer

Discussion

3040.	\frac { 5 ^ { n + 2 } - 5 ^ { n + 1 } } { 5 ^ { n + 3 } }
Answer» Thank u yaar

Discussion

3041.	27. If 0-30", verify the following:i) Cos 30-4 cos' -3 cose,ii) Sin 30-3 sin -4sin'
Answer»

Discussion

3042.	30 ^ { 3 } + 20 ^ { 3 } - 50 ^ { 3 } + 90000
Answer»

Discussion

3043.	The square of the diagonal of a square is 98 sq cm. Find the side of a square.
Answer»

Discussion

3044.	length of the diagonal of a square is 6v2 cm. Fin its area.
Answer»

Discussion

3045.	3)-14) 2If x tan 45° sin 30° cos 30° tan 30°, then x is equalto1(3)石(1)32(91(2)2
Answer» xtan45sin30 =cos30 tan 30x (1) (1/2)=(sqrt(3)/2) (1/sqrt(3))x=1 pleace Complate answer

3045.

3)-14) 2If x tan 45° sin 30° cos 30° tan 30°, then x is equalto1(3)石(1)32(91(2)2

Answer»

xtan45sin30 =cos30 tan 30x (1) (1/2)=(sqrt(3)/2) (1/sqrt(3))x=1

pleace Complate answer

Discussion

3046.	(1) 454) 30(3) 6:3:2
Answer» option (c) is correct.

Discussion

3047.	0EXERCISE 11E1. Find the value of the following.i. 117 x V157315.29 × 13.69
Answer» Second wala please

Discussion

3048.	square %2B 8=6
Answer» -2 the correct answer of the given question -2................................................................................... -2 is the answer of this most dangerous question -2 is right answer.... -2 is the correct answer of the given question - 2 the correct answer of your question correct answer is (-2) -2 is the correct answer

3048.

square %2B 8=6

Answer»

-2 the correct answer of the given question

-2...................................................................................

-2 is the answer of this most dangerous question

-2 is right answer....

-2 is the correct answer of the given question

- 2 the correct answer of your question

correct answer is (-2)

-2 is the correct answer

Discussion

3049.	Without actual division, show that (3-3x2-13x +15) is exactlydivisible by (a2+2x-3).
Answer»

Discussion

3050.	6. Find the value of a so that a+x-4x^2+x^3 is exactlydivisible by x - 3.
Answer» divisible by x-3 so if we put x=3 equation will be equal to 0a+3-4(9)+(3)^3=0a+3-36+27=0so a=6

Discussion

Explore topic-wise InterviewSolutions in .

Two cylindrical vessels are filled with milk. The radius of one vessel is 15 cm andheight is 40 cm, and the radius of other vessel is 20 cm and height is 45 cm. Find theradius of another cylindrical vessel of height 30 em which may just contain the milkwhich is in the two given vessels.

areaBase18 cmHeight15cm

.उदाहरण 15. सिद्ध कीजिए कि:ca 30.

A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. Thisucket is emptied on the ground and a conical heap of sand is formed. If the height of theconical heap is 24 cm, find the radius and slant height of the heap.

A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand Tbucket is emptied on the ground and a conical heap of sand is formedconical heap is 24 cm, find the radius and slant height of the heap.If the height of's

7. Acylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. Thisbucket is emptied on the ground and a conical heap of sand is formed. Ifthe height of theconical heap is 24 cm, find the radius and slant height of the heap.

A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. Thisbucket is emptied on the ground and a conical heap of sand is formed. If the height of theconical heap is 24 cm, find the radius and slant height of the heap.

7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled withand. Thibucket is emptied on the ground and a conical heap of sand is formed. If the height of theconical heap is 24 cm, find the radius and slant height of the heap.

20. Determine the missing frequency x, from thefollowing data, when mode is 67.Class40-50 50-60 -60-70 70-80 80-90Frequency5x 15127

7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filledThnd is formed. If the height of thebucket is emptied on the ground and a conical heap of sand is formed. Ifthconical heap is 24 cm, find the radius and slant height of the heap

The Class marks of a distribution are 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97, and 102.Determine the Class size?

5.The Class marks of a distribution are 47,52, 57, 62, 67, 72, 77, 82, 87, 92, 97, and 102.Determine the Class size?

x2+ 18x +81

(7*t %2B 2)/(-5*t %2B 1)=(-7*t %2B 3)/(5*t %2B 4)

(a)18x(7 + (-3)] = [18×7] + [ 18 × (-3)]

e oooके=0 z ¢v o Yo c"* !«}3"45 .QY पी हिa8 Qe: QN _‘j Nहि १(दी १|)

4x^4+x^3-73x^2-18x divide by x^2-18

) 18 рек [7 + (-3)] * [18 * 7] + [18 x (-3)]..... 4 MBS &amp;

18x+18x=72 then x=?

Obtain other zeroes of the polynomial 4xx-x-18x, if two of its zeroes are 32 and-3/2

18x=13x+62

(21*(text*(18*(a*(d*n)))))*(-3 %2B 21) %2B 18*(-3)

7. If θ(i)30°, verify thatsin 2θ2 sin θ cos θ

If $ \theta=30^{\circ}, $ verify that$ 1-\cos 2 \theta=2 \sin ^{2} \theta $

1/13 %2B 3*sqrt(3) %2B 2*(7*sqrt(2))

40.Current I in the circuit will be :-20Ω30 Ω20 Ω5 V5(1)a A(2) 50 A(3)to AOA(4)

a) 65x13

\frac { 4 \sqrt { 3 } } { 2 - \sqrt { 2 } } - \frac { 30 } { 4 \sqrt { 3 } - \sqrt { 18 } } - \frac { \sqrt { 18 } } { 3 - \sqrt { 12 } }

\frac{4 \sqrt{3}}{2-\sqrt{2}}-\frac{30}{4 \sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}

+ bv/3, then find a and(3)If (5W2 + 3:3) _ (6V2-7:3) = a12

8. Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.

30525 aI8\frac{4 \sqrt{3}}{2-\sqrt{2}}-\frac{30}{4 \sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}

2. Solve;(c) 77318 18 (15 15+3 (g)3

If $ ^{18} C_{15}+2^{18} C_{16}+17 C_{16}+1=^{n} C_{3} $ then(1) 19(3) 18

(a) 18 x [7+ (-3)] [18 x 7]+ [18x (-3)]

\frac { 5 ^ { n + 2 } - 6 \times 5 ^ { n + 1 } } { 13 \times 5 ^ { n } - 2 \times 5 ^ { n + 1 } } Q

\frac { 5 ^ { n + 2 } - 6 \times 5 ^ { n + 1 } } { 13 \times 5 ^ { n } - 2 \times 5 ^ { n + 1 } }

\frac{10 \times 5^{n+1}+25 \times 5^{n}}{3 \times 5^{n+2}+10 \times 5^{n+1}}

(-6*5^(n %2B 1) %2B 5^(n %2B 2))/(13*5^n - 2*5^(n %2B 1))

\frac { 5 ^ { n + 2 } - 5 ^ { n + 1 } } { 5 ^ { n + 3 } }

27. If 0-30", verify the following:i) Cos 30-4 cos' -3 cose,ii) Sin 30-3 sin -4sin'

30 ^ { 3 } + 20 ^ { 3 } - 50 ^ { 3 } + 90000

The square of the diagonal of a square is 98 sq cm. Find the side of a square.

length of the diagonal of a square is 6v2 cm. Fin its area.

3)-14) 2If x tan 45° sin 30° cos 30° tan 30°, then x is equalto1(3)石(1)32(91(2)2

(1) 454) 30(3) 6:3:2

0EXERCISE 11E1. Find the value of the following.i. 117 x V157315.29 × 13.69

square %2B 8=6

Without actual division, show that (3-3x2-13x +15) is exactlydivisible by (a2+2x-3).

6. Find the value of a so that a+x-4x^2+x^3 is exactlydivisible by x - 3.

(7t %2B 2)/(-5t %2B 1)=(-7t %2B 3)/(5t %2B 4)

) 18 рек [7 + (-3)] * [18 * 7] + [18 x (-3)]..... 4 MBS &

(21(text(18(a(dn)))))(-3 %2B 21) %2B 18*(-3)

1/13 %2B 3sqrt(3) %2B 2(7*sqrt(2))

(-65^(n %2B 1) %2B 5^(n %2B 2))/(135^n - 2*5^(n %2B 1))