332354 + Interview Questions in Secondary School in Math

1.	out of men and 3women ,a committe of three members is to be formed so that it has 1 women ,and 2men .in how many diffrent ways can it be done
Answer» ANSWER: ??????????????? ¿?????? ¿?????????.????.. Step-by-step EXPLANATION: ??????????????????????

1.

out of men and 3women ,a committe of three members is to be formed so that it has 1 women ,and 2men .in how many diffrent ways can it be done

Answer»

ANSWER:

??????????????? ¿?????? ¿?????????.????..

Step-by-step EXPLANATION:

??????????????????????

Discussion

2.	Higest marks-lowest marks
Answer» Higest MARKS : 99 lowest marks :38

2.

Higest marks-lowest marks

Answer»

Higest MARKS : 99

lowest marks :38

Discussion

3.	Compare √5 upon 3 ,3 upon √7
Answer» Step-by-step EXPLANATION: What do you EXACTLY MEAN by COMPARE?

3.

Compare √5 upon 3 ,3 upon √7

Answer»

Step-by-step EXPLANATION:

What do you EXACTLY MEAN by COMPARE?

Discussion

4.	गिफ्ट वैल्यू ऑफ वन जीरो इन वन मोर देन द अदर जीरो पॉलिनॉमियल पी ऑफ एक्स इन एक्स स्क्वायर प्लस 5 एक्स प्लस के 10 फाइंड द वैल्यू ऑफ के
Answer» Answer: 1111111 Step-by-step explanation: 44444444&&+9-+9&

4.

गिफ्ट वैल्यू ऑफ वन जीरो इन वन मोर देन द अदर जीरो पॉलिनॉमियल पी ऑफ एक्स इन एक्स स्क्वायर प्लस 5 एक्स प्लस के 10 फाइंड द वैल्यू ऑफ के

Answer»

Answer:

1111111

Step-by-step explanation:

44444444&&+9-+9&

Discussion

5.	The factors of 36x2 – 25 are
Answer» Step-by-step explanation: ALGEBRA Examples Rewrite 36x2 36 x 2 as (6x)2 ( 6 x ) 2 . Rewrite 25 as 52 . SINCE both terms are PERFECT squares, factor USING the difference of squares formula, a2−b2=(a+b)(a−b) a 2 - b 2 = ( a + b ) ( a - b ) where a=6x a = 6 x and b=5 .

5.

The factors of 36x2 – 25 are

Answer»

Step-by-step explanation:

ALGEBRA Examples

Rewrite 36x2 36 x 2 as (6x)2 ( 6 x ) 2 . Rewrite 25 as 52 . SINCE both terms are PERFECT squares, factor USING the difference of squares formula, a2−b2=(a+b)(a−b) a 2 - b 2 = ( a + b ) ( a - b ) where a=6x a = 6 x and b=5 .

Discussion

6.	+++Convert ento decimal fracking 500g
Answer» Step-by-step EXPLANATION: .fyfcuvyfngjuyyyyyyyyyyyyyyyy

6.

+++Convert ento decimal fracking 500g

Answer»

Step-by-step EXPLANATION:

.fyfcuvyfngjuyyyyyyyyyyyyyyyy

Discussion

7.	In the given figures, measure of some parts of the triangle are indicated.By applying SAS congruence rule.State the pair of congruent triangles, if any.In case of congruent triangles, write them in symbolic form.
Answer» ANSWER: GOOD Step-by-step EXPLANATION: good good good

7.

In the given figures, measure of some parts of the triangle are indicated.By applying SAS congruence rule.State the pair of congruent triangles, if any.In case of congruent triangles, write them in symbolic form.

Answer»

ANSWER:

GOOD

Step-by-step EXPLANATION:

good good good

Discussion

8.	A boy runs 1 km in 10 minutes. How long will he take to ran 600m?
Answer» Answer: 600 hunsred MINUTES it will TAKE

8.

A boy runs 1 km in 10 minutes. How long will he take to ran 600m?

Answer»

Answer:

600 hunsred MINUTES it will TAKE

Discussion

9.	Find the difference between 31m and 8 m 86 cm
Answer» Answer: 86cm=0.86m so,8m86cm=8.86m Thus,DIFFERENCE between 31M and 8.86m=(31-8.86)m=22.14m

9.

Find the difference between 31m and 8 m 86 cm

Answer»

Answer:

86cm=0.86m

so,8m86cm=8.86m

Thus,DIFFERENCE between 31M and 8.86m=(31-8.86)m=22.14m

Discussion

10.	Subtract 10xy from 14xy.
Answer» Step-by-step EXPLANATION: Here, 10xy, xy are like terms The difference of the COEFFICIENTS = 10 - 1, [xy MEANS 1xy] Therefore, 10xy - xy = 9xy

10.

Subtract 10xy from 14xy.

Answer»

Step-by-step EXPLANATION:

Here, 10xy, xy are like terms

The difference of the COEFFICIENTS = 10 - 1, [xy MEANS 1xy]

Therefore, 10xy - xy = 9xy

Discussion

11.	Find the quadratic polynomial whose one zero is 3and product of zero is 12
Answer» Step-by-step EXPLANATION: e9odrofof9r8rir8rueur737e7e7374748r83848r8

11.

Find the quadratic polynomial whose one zero is 3and product of zero is 12

Answer»

Step-by-step EXPLANATION:

e9odrofof9r8rir8rueur737e7e7374748r83848r8

Discussion

12.	Find the area of rectangle whose length is 10 cm and breath 50 cm
Answer» ✭ CORRECT Question : Find the area of rectangle whose length is 10 cm and breath 50 cm ✭ Required SOLUTION : ✦ FORMULA used here : Area = length × Breath ✭ Value given to us : Length (l) = 10 cm Breath (B) = 50 cm ✭ Putting values in formula : A = 10 × 50 cm A = 500 cm² ✰ Area of Rectangle is 500 cm² ✰ ___________________________

12.

Find the area of rectangle whose length is 10 cm and breath 50 cm

Answer»

✭ CORRECT Question :

Find the area of rectangle whose length is 10 cm and breath 50 cm

✭ Required SOLUTION :

✦ FORMULA used here :

Area = length × Breath

✭ Value given to us :

Length (l) = 10 cm

Breath (B) = 50 cm

✭ Putting values in formula :

A = 10 × 50 cm

A = 500 cm²

✰ Area of Rectangle is 500 cm² ✰

___________________________

Discussion

13.	55-11 [120:24 +10+5-7)
Answer» ANSWER: , Step-by-step EXPLANATION: THATS your answer you have to SOLVE 1372/24 if they both are DIVIDED then you have 57.6

13.

55-11 [120:24 +10+5-7)

Answer»

ANSWER:

,

Step-by-step EXPLANATION:

THATS your answer you have to SOLVE 1372/24 if they both are DIVIDED then you have 57.6

Discussion

14.	Ram bought a second hand cycle for 500. He spent 80 in repairing and175 in repainting. He then sold it to Lalit for 900. How much did he gain orlose?
Answer» Answer: 500+80=180 180+175=355 then he SOLD it for 900 900-335=545 so he has a PROFIT of Rs545

14.

Ram bought a second hand cycle for 500. He spent 80 in repairing and175 in repainting. He then sold it to Lalit for 900. How much did he gain orlose?

Answer»

Answer:

500+80=180

180+175=355

then he SOLD it for 900

900-335=545

so he has a PROFIT of Rs545

Discussion

15.	Differentiate the function w. r. t x e^x / 1+x^2
Answer» Let y= (X−3)(x−4)(x−5) (x−1)(x−2) TAKING logarithm on both the sides, we obtain logy=log (x−3)(x−4)(x−5) (x−1)(x−2) ⇒logy= 2 1 log[ (x−3)(x−4)(x−5) (x−1)(x−2) ] ⇒logy= 2 1 [log[(x−1)(x−2)]−log[(x−3)(x−4)(x−5)]] ⇒logy= 2 1 [log(x−1)+log(x−2)−log(x−3)−log(x−4)−log(x−5)] Differentiating both sides with respect to x, we obtain ⇒ y 1 dx dy = 2 1 [ x−1 1 .+ x−2 1 − x−3 1 − x−4 1 − x−5 1 ] ⇒ dx dy = 2 y ( x−1 1 + x−2 1 − x−3 1 − x−4 1 − x−5 1 ) ∴ dx dy = 2 1 (x−3)(x−4)(x−5) (x−1)(x−2) [ x−1 1 + x−2 1 − x−3 1 − x−4 1 − x−5 1 ]

15.

Differentiate the function w. r. t x e^x / 1+x^2

Answer»

Let y=

(X−3)(x−4)(x−5)

(x−1)(x−2)

TAKING logarithm on both the sides, we obtain

logy=log

(x−3)(x−4)(x−5)

(x−1)(x−2)

⇒logy=

2

1 log[

(x−3)(x−4)(x−5)

(x−1)(x−2)

]

⇒logy=

2

1 [log[(x−1)(x−2)]−log[(x−3)(x−4)(x−5)]]

⇒logy=

2

1 [log(x−1)+log(x−2)−log(x−3)−log(x−4)−log(x−5)]

Differentiating both sides with respect to x, we obtain

⇒

y

1 dx

dy

=

2

1 [

x−1

1 .+

x−2

1 −

x−3

1 −

x−4

1 −

x−5

1 ]

⇒

dx

dy

=

2 y

(

x−1

1 +

x−2

1 −

x−3

1 −

x−4

1 −

x−5

1 )

∴

dx

dy

=

2

1 (x−3)(x−4)(x−5)

(x−1)(x−2)

[

x−1

1 +

x−2

1 −

x−3

1 −

x−4

1 −

x−5

1 ]

Discussion

16.	A ornament weighs 45g.It is a mixture of gold and copper in the ratio 7:2.The weight of copper is______.
Answer» Answer: 10g Step-by-step explanation: let wt of GOLD and copper be 7X and 9X then 7x+2x = 45g hence X= 5g hence wt if copper=2x=10g

16.

A ornament weighs 45g.It is a mixture of gold and copper in the ratio 7:2.The weight of copper is______.

Answer»

Answer:

10g

Step-by-step explanation:

let wt of GOLD and copper be 7X and 9X

then 7x+2x = 45g

hence X= 5g

hence wt if copper=2x=10g

Discussion

17.	(xdx -ydy) y sin y/x = ( ydx + xdx) x cos y/x
Answer» Correct QUESTION:- SOLVE the differential EQUATION $(x\ dy-y\ dx)y\sin\left(\dfrac{y}{x}\right)=(y\ dx+x\ dy)x\cos\left(\dfrac{y}{x}\right).$ Solution:- We're GIVEN to solve the differential equation, $\longrightarrow(x\ dy-y\ dx)y\sin\left(\dfrac{y}{x}\right)=(y\ dx+x\ dy)x\cos\left(\dfrac{y}{x}\right)$ Dividing by $x^2$ we get, $\longrightarrow\left(\dfrac{x\ dy-y\ dx}{x^2}\right)y\sin\left(\dfrac{y}{x}\right)=\left(\dfrac{y}{x}\ dx+dy\right)\cos\left(\dfrac{y}{x}\right)$ Put, $\longrightarrow u=\dfrac{y}{x}$ $\longrightarrow du=\dfrac{x\ dy-y\ dx}{x^2}$ Also, $\longrightarrow y=ux$ $\longrightarrow dy=u\ dx+x\ du$ Then our equation BECOMES, $\longrightarrow ux\sin u\ du=\left(2u\ dx+x\ du\right)\cos u$ $\longrightarrow ux\sin u\ du=2u\cos u\ dx+x\cos u\ du$ $\longrightarrow ux\sin u\ du-x\cos u\ du=2u\cos u\ dx$ $\longrightarrow(u\sin u-\cos u)x\ du=2u\cos u\ dx$ $\longrightarrow\dfrac{dx}{x}=\left(\dfrac{u\sin u-\cos u}{2u\cos u}\right)\ du$ $\longrightarrow\dfrac{dx}{x}=\dfrac{1}{2}\left(\tan u-\dfrac{1}{u}\right)\ du$ Now integrating, $\displaystyle\longrightarrow\int\dfrac{dx}{x}=\dfrac{1}{2}\int\left(\tan u-\dfrac{1}{u}\right)\ du$ $\displaystyle\longrightarrow\log\|x\|=\dfrac{1}{2}\left(\log\|\sec u\|-\log\|u\|\right)+\log\|C_1\|$ $\displaystyle\longrightarrow\log\|x\|=\dfrac{1}{2}\log\left\|\dfrac{\sec u}{u}\right\|+\log\|C_1\|$ Taking antilog, $\displaystyle\longrightarrow x=C_1\sqrt{\dfrac{\sec u}{u}}$ $\displaystyle\longrightarrow x^2=\dfrac{(C_1)^2\sec u}{u}$ $\displaystyle\longrightarrow \sec u=\dfrac{ux^2}{(C_1)^2}$ Taking $(C_1)^2=\dfrac{1}{C},$ $\displaystyle\longrightarrow \sec u=Cux^2$ Undoing substitution $u=\dfrac{y}{x},$ $\displaystyle\longrightarrow\underline{\underline{\sec\left(\dfrac{y}{x}\right)=Cxy}}$

17.

(xdx -ydy) y sin y/x = ( ydx + xdx) x cos y/x

Answer»

Correct QUESTION:-

SOLVE the differential EQUATION $(x\ dy-y\ dx)y\sin\left(\dfrac{y}{x}\right)=(y\ dx+x\ dy)x\cos\left(\dfrac{y}{x}\right).$

Solution:-

We're GIVEN to solve the differential equation,

$\longrightarrow(x\ dy-y\ dx)y\sin\left(\dfrac{y}{x}\right)=(y\ dx+x\ dy)x\cos\left(\dfrac{y}{x}\right)$

Dividing by $x^2$ we get,

$\longrightarrow\left(\dfrac{x\ dy-y\ dx}{x^2}\right)y\sin\left(\dfrac{y}{x}\right)=\left(\dfrac{y}{x}\ dx+dy\right)\cos\left(\dfrac{y}{x}\right)$

Put,

$\longrightarrow u=\dfrac{y}{x}$

$\longrightarrow du=\dfrac{x\ dy-y\ dx}{x^2}$

Also,

$\longrightarrow y=ux$

$\longrightarrow dy=u\ dx+x\ du$

Then our equation BECOMES,

$\longrightarrow ux\sin u\ du=\left(2u\ dx+x\ du\right)\cos u$

$\longrightarrow ux\sin u\ du=2u\cos u\ dx+x\cos u\ du$

$\longrightarrow ux\sin u\ du-x\cos u\ du=2u\cos u\ dx$

$\longrightarrow(u\sin u-\cos u)x\ du=2u\cos u\ dx$

$\longrightarrow\dfrac{dx}{x}=\left(\dfrac{u\sin u-\cos u}{2u\cos u}\right)\ du$

$\longrightarrow\dfrac{dx}{x}=\dfrac{1}{2}\left(\tan u-\dfrac{1}{u}\right)\ du$

Now integrating,

$\displaystyle\longrightarrow\int\dfrac{dx}{x}=\dfrac{1}{2}\int\left(\tan u-\dfrac{1}{u}\right)\ du$

$\displaystyle\longrightarrow\log|x|=\dfrac{1}{2}\left(\log|\sec u|-\log|u|\right)+\log|C_1|$

$\displaystyle\longrightarrow\log|x|=\dfrac{1}{2}\log\left|\dfrac{\sec u}{u}\right|+\log|C_1|$

Taking antilog,

$\displaystyle\longrightarrow x=C_1\sqrt{\dfrac{\sec u}{u}}$

$\displaystyle\longrightarrow x^2=\dfrac{(C_1)^2\sec u}{u}$

$\displaystyle\longrightarrow \sec u=\dfrac{ux^2}{(C_1)^2}$

Taking $(C_1)^2=\dfrac{1}{C},$

$\displaystyle\longrightarrow \sec u=Cux^2$

Undoing substitution $u=\dfrac{y}{x},$

$\displaystyle\longrightarrow\underline{\underline{\sec\left(\dfrac{y}{x}\right)=Cxy}}$

Discussion

18.	प्रश्न.12 एक आयताकार भूखण्ड की लम्बाई और चौड़ाई क्रमशः 500 मीटरतथा 300 मीटर है । तो भूखण्ड का क्षेत्रफल ज्ञात कीजिए ।
Answer» Step-by-step EXPLANATION: bhookhand KA kshetrafal = LAMBAI × chaudai so , 500× 300 = 150000 m²

18.

प्रश्न.12 एक आयताकार भूखण्ड की लम्बाई और चौड़ाई क्रमशः 500 मीटरतथा 300 मीटर है । तो भूखण्ड का क्षेत्रफल ज्ञात कीजिए ।

Answer»

Step-by-step EXPLANATION:

bhookhand KA kshetrafal = LAMBAI × chaudai

so , 500× 300 = 150000 m²

Discussion

19.	2.Draw a figure representing fraction : 5_8ת \| Co
Answer» $\huge{5/8}$

19.

2.Draw a figure representing fraction : 5_8ת | Co

Answer»

$\huge{5/8}$

Discussion

20.	Express the centimeter and kilometre in meter
Answer» Answer: 1cm=1/100 METERS 1km=1000 meters

20.

Express the centimeter and kilometre in meter

Answer»

Answer:

1cm=1/100 METERS

1km=1000 meters

Discussion

21.	Tangent theorem solution
Answer» Answer: If a tangent segment and a secant segment are DRAWN to a CIRCLE from an exterior point, then the SQUARE of the measure of the tangent segment is equal to the product of the MEASURES of the secant segment and its external secant segment. In the circle, ¯UV is a tangent and ¯UY is a secant. They intersect at point U .

21.

Tangent theorem solution

Answer»

Answer:

If a tangent segment and a secant segment are DRAWN to a CIRCLE from an exterior point, then the SQUARE of the measure of the tangent segment is equal to the product of the MEASURES of the secant segment and its external secant segment. In the circle, ¯UV is a tangent and ¯UY is a secant. They intersect at point U .

Discussion

22.	3/5+6/7+4/8 add the following fraction.
Answer» ANSWER: ANS= 1.95714285714 CORRECT answer FOLLOW

22.

3/5+6/7+4/8 add the following fraction.

Answer»

ANSWER:

ANS= 1.95714285714 CORRECT answer FOLLOW

Discussion

23.	The ages of three men are in the ratio 3:4:5. If the difference between the ages of the oldest and youngest is 18 years, find the sum of the ages of the three men.
Answer» Let the ages be 3x, 4X and 5X years. 5x - 3x = 18 2x = 18 x = 9 5x + 4x + 3x 12x 12 × 9 108 The sum of their ages is 108 years.

23.

The ages of three men are in the ratio 3:4:5. If the difference between the ages of the oldest and youngest is 18 years, find the sum of the ages of the three men.

Answer»

Let the ages be 3x, 4X and 5X years.

5x - 3x = 18

2x = 18

x = 9

5x + 4x + 3x

12x

12 × 9

108

The sum of their ages is 108 years.

Discussion

24.	Find the value using distributive property(-65) 25 + 25x (-35)
Answer» ANSWER Solution : (-65) x 25 + 25 x (-35) 25 [(-65) + (-35)] => 25 [-65 -35] => 25 (-100) => - 2500 Hence, the required answer is - 2500.

24.

Find the value using distributive property(-65) 25 + 25x (-35)

Answer»

ANSWER

Solution :

(-65) x 25 + 25 x (-35)

25 [(-65) + (-35)]

=> 25 [-65 -35]

=> 25 (-100)

=> - 2500

Hence, the required answer is - 2500.

Discussion

25.	Find the value of x in the equation x+5=8
Answer» Step-by-step EXPLANATION: x+5 = 8 x= 8-5 x= 3 I HOPE it HELPS you

25.

Find the value of x in the equation x+5=8

Answer»

Step-by-step EXPLANATION:

x+5 = 8

x= 8-5

x= 3

I HOPE it HELPS you

Discussion

26.	Without actual division, find which of the following fractions are terminating decimals :-(iv) 25/128Please show the full process!
Answer» $\huge✎\fbox \orange{QUE} {\colorbox{blue}{ST}}\fbox\green{ION}☟$ $\orange{\Large{\mathcal\colorbox{indigo}{Is 25/128 a Terminating Decimal}}}$ $\huge✍︎\fbox\orange{ÂŇ}\fbox{SW} \fbox\green{ÊŘ}:$ Brother/Sister your CORRECT answer will be:- $\purple{\huge{\mathcal\colorbox{pink}{Yes}}}$ Step-by-step EXPLANATION:- $\purple{The \: given \: number \: is} \: \orange{\boxed{\red{\frac{25}{128}}}}$ $\blue{Its \: HCF(25,128) = 1}$ $\therefore {{\purple{\boxed{\red{\frac{25}{128}}}}} \: {is \: in \: simplest \: form.}}$ $\red{Now, \: 128 = 2 \times 2 \times 2 \times 2 \times2 \times 2 \times2 \:[Prime \: Factorization]}$ $\orange{So, \: its \: prime \: factor \: is \: 2}$ $\green{Hence, \: {\orange{\boxed{ \red{\frac{25}{128}}}}} \: is \: a \: terminating \: decimal.}$ $\pink{꧁Hopes\:so\:that\:this\:will\:help\:you,꧂}$ $\huge\colorbox{orange}❤\colorbox{indigo}{\red{With\:Love}}\colorbox{orange}❤$ ✌︎Thanking you, your one of the Brother of your 139 Crore INDIAN BROTHERS and Sisters.✌︎ $\large\red{\underbrace{\overbrace\mathbb{\colorbox{aqua}{\colorbox{blue}{\colorbox{purple}{\colorbox{pink}{\colorbox{green}{\colorbox{gold}{\colorbox{silver}{\textit{\purple{\underline{\red{《❀PŘÎ᭄ŇCĒ✿》࿐}}}}}}}}}}}}}}$

26.

Without actual division, find which of the following fractions are terminating decimals :-(iv) 25/128Please show the full process!

Answer»

$\huge✎\fbox \orange{QUE} {\colorbox{blue}{ST}}\fbox\green{ION}☟$

$\orange{\Large{\mathcal\colorbox{indigo}{Is 25/128 a Terminating Decimal}}}$

$\huge✍︎\fbox\orange{ÂŇ}\fbox{SW} \fbox\green{ÊŘ}:$

Brother/Sister your CORRECT answer will be:-

$\purple{\huge{\mathcal\colorbox{pink}{Yes}}}$

Step-by-step EXPLANATION:-

$\purple{The \: given \: number \: is} \: \orange{\boxed{\red{\frac{25}{128}}}}$

$\blue{Its \: HCF(25,128) = 1}$

$\therefore {{\purple{\boxed{\red{\frac{25}{128}}}}} \: {is \: in \: simplest \: form.}}$

$\red{Now, \: 128 = 2 \times 2 \times 2 \times 2 \times2 \times 2 \times2 \:[Prime \: Factorization]}$

$\orange{So, \: its \: prime \: factor \: is \: 2}$

$\green{Hence, \: {\orange{\boxed{ \red{\frac{25}{128}}}}} \: is \: a \: terminating \: decimal.}$

$\pink{꧁Hopes\:so\:that\:this\:will\:help\:you,꧂}$

$\huge\colorbox{orange}❤\colorbox{indigo}{\red{With\:Love}}\colorbox{orange}❤$

✌︎Thanking you, your one of the Brother of your 139 Crore INDIAN BROTHERS and Sisters.✌︎

$\large\red{\underbrace{\overbrace\mathbb{\colorbox{aqua}{\colorbox{blue}{\colorbox{purple}{\colorbox{pink}{\colorbox{green}{\colorbox{gold}{\colorbox{silver}{\textit{\purple{\underline{\red{《❀PŘÎ᭄ŇCĒ✿》࿐}}}}}}}}}}}}}}$

Discussion

27.	33. ₹18440 पर 15% की वार्षिक दर से 4 वर्ष का साधारण ब्याज कितना होगा?(a)₹11064 (b)₹ 12250 (c) ₹11500(d) ₹11075
Answer» ANSWER: ₹18440 पर 15% की वार्षिक दर से 4 वर्ष का साधारण ब्याज कितना होगा? (a)₹11064 (B) ₹12250 (C) ₹11500 (d) ₹11075. Answer 18440 × 15/100 = 12250

27.

33. ₹18440 पर 15% की वार्षिक दर से 4 वर्ष का साधारण ब्याज कितना होगा?(a)₹11064 (b)₹ 12250 (c) ₹11500(d) ₹11075

Answer»

ANSWER:

₹18440 पर 15% की वार्षिक दर से 4 वर्ष का साधारण ब्याज कितना होगा? (a)₹11064 (B) ₹12250 (C) ₹11500 (d) ₹11075.

Answer 18440 × 15/100 = 12250

Discussion

28.	Write the number of 10 paise coins 18.00 full explanation step by step
Answer» Answer: Number of 10 PAISE coin in one RUPEE = 100 ÷ 10 = 10 TEN 10 paise coin will make one rupee.

28.

Write the number of 10 paise coins 18.00 full explanation step by step

Answer»

Answer:

Number of 10 PAISE coin in one RUPEE = 100 ÷ 10 = 10 TEN 10 paise coin will make one rupee.

Discussion

29.	Write full form of GST
Answer» ANSWER: GST Full FORM is Goods and Services Tax. Step-by-step explanation: GST has been introduced to replace multiple INDIRECT taxes levied by STATE and Central Governments in order to simplify the indirect tax system. ... It is CALLED as Goods and Services Tax because it is applicable on the supply of both Goods and Services.

29.

Write full form of GST

Answer»

ANSWER:

GST Full FORM is Goods and Services Tax.

Step-by-step explanation:

GST has been introduced to replace multiple INDIRECT taxes levied by STATE and Central Governments in order to simplify the indirect tax system. ... It is CALLED as Goods and Services Tax because it is applicable on the supply of both Goods and Services.

Discussion

30.	Find the value of x when x tanπ/4.sinπ/3.cotπ/6= sec 0ºcosec 30°
Answer» ANSWER: π/4=45° ; π/3=60° ; π/6=30° xtan45°.sin60°.cot30°=sec0°·cosec30° x(1).[√3]/2.√3=(1)2 x(3/2)=2 x=4/3 :)

30.

Find the value of x when x tanπ/4.sinπ/3.cotπ/6= sec 0ºcosec 30°

Answer»

ANSWER:

π/4=45° ; π/3=60° ; π/6=30°

xtan45°.sin60°.cot30°=sec0°·cosec30°

x(1).[√3]/2.√3=(1)2

x(3/2)=2

x=4/3

:)

Discussion

31.	1112Fig. 10.5if angle 1 is55° find angle
Answer» Answer: 55° Step-by-step EXPLANATION: angle 1 = angle 2 = 55° Reason : CORRESPONDING angles are equal.

31.

1112Fig. 10.5if angle 1 is55° find angle

Answer»

Answer:

55°

Step-by-step EXPLANATION:

angle 1 = angle 2 = 55°

Reason : CORRESPONDING angles are equal.

Discussion

32.	If the volume of an cylinder is 700 cubic centimeters and the area of the base is 100 square centimeters, what is the length of the cylinder?*1️⃣ 70 cm2️⃣ 7 cm3️⃣ 700 cm4️⃣ 7 sqcm
Answer» Given : Volume of an CYLINDER is 700 cm³ and the AREA of the base is 100 cm² . Need To Find ; The Length or HEIGHT of Cylinder . ⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ ❍ Let's CONSIDER the Length or Height of Cylinder is h cm . $\dag\frak{\underline { As,\;We\:know\:that \::}}\\$ $\star\boxed {\pink{\sf{ Curved \:Surface \:area\:_{(Cylinder)} = \pi r^{2} h }}}\\\\$ And , $\star\boxed {\pink{\sf{ \:Area\:_{(Circle)} = \pi r^{2} }}}\\\\$ Where , r is the Radius of Circle and Cylinder & h is the Height or Length of Cylinder & $\pi =\dfrac{22}{7}$ Given that , Area of Base Circle of Cylinder is 100 cm² Or , $\pi$ r² = 100 cm² ⠀⠀⠀⠀⠀⠀ $\underline {\bf{\star\:Now \: By \: Substituting \: the \: known\: Values \::}}\\$ $\qquad \quad :\implies\sf{ 700 = \pi r^{2} h }\\\\$ $\pi$ r² = 100 cm² $\qquad \quad :\implies\sf{ 700 = 100( h) }\\\\$ $\qquad \quad :\implies\sf{ \cancel {\dfrac{700}{100}} = h }\\\\$ ⠀⠀⠀⠀⠀ $\underline {\boxed{\pink{ \mathrm { h = 7\: cm}}}}\:\bf{\bigstar}\\$ Therefore, ⠀⠀⠀⠀⠀ $\therefore {\underline{ \mathrm { Hence,\:Length \:of\:Cylinder \:is\:\bf{7\: cm}}}}\\$ ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

32.

If the volume of an cylinder is 700 cubic centimeters and the area of the base is 100 square centimeters, what is the length of the cylinder?*1️⃣ 70 cm2️⃣ 7 cm3️⃣ 700 cm4️⃣ 7 sqcm

Answer»

Given : Volume of an CYLINDER is 700 cm³ and the AREA of the base is 100 cm² .

Need To Find ; The Length or HEIGHT of Cylinder .

⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Let's CONSIDER the Length or Height of Cylinder is h cm .

$\dag\frak{\underline { As,\;We\:know\:that \::}}\\$

$\star\boxed {\pink{\sf{ Curved \:Surface \:area\:_{(Cylinder)} = \pi r^{2} h }}}\\\\$

And ,

$\star\boxed {\pink{\sf{ \:Area\:_{(Circle)} = \pi r^{2} }}}\\\\$

Where ,

r is the Radius of Circle and Cylinder & h is the Height or Length of Cylinder & $\pi =\dfrac{22}{7}$

Given that ,

Area of Base Circle of Cylinder is 100 cm²

Or ,

$\pi$ r² = 100 cm²

⠀⠀⠀⠀⠀⠀ $\underline {\bf{\star\:Now \: By \: Substituting \: the \: known\: Values \::}}\\$

$\qquad \quad :\implies\sf{ 700 = \pi r^{2} h }\\\\$

$\pi$ r² = 100 cm²

$\qquad \quad :\implies\sf{ 700 = 100( h) }\\\\$

$\qquad \quad :\implies\sf{ \cancel {\dfrac{700}{100}} = h }\\\\$

⠀⠀⠀⠀⠀ $\underline {\boxed{\pink{ \mathrm { h = 7\: cm}}}}\:\bf{\bigstar}\\$

Therefore,

⠀⠀⠀⠀⠀ $\therefore {\underline{ \mathrm { Hence,\:Length \:of\:Cylinder \:is\:\bf{7\: cm}}}}\\$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Discussion

33.	5x+2y=16,3x+6y/5=2 find the type of solution
Answer» Answer: n engl to laga lijiye the hi h to FIR se bat the valve and U to laga lijiye Janam Diya woh to the kal chalegaye and night the valve is the hi NHI mil pata nahi kya KAR do PK sanny the valve means of a good time for 47 to the kal chalegaye

33.

5x+2y=16,3x+6y/5=2 find the type of solution

Answer»

Answer:

n engl to laga lijiye the hi h to FIR se bat the valve and U to laga lijiye Janam Diya woh to the kal chalegaye and night the valve is the hi NHI mil pata nahi kya KAR do PK sanny the valve means of a good time for 47 to the kal chalegaye

Discussion

34.	Find two rational numbers between ½ and 5/4.
Answer» -1 / 4,0,1 / 4,2 / 4,3 / 4 and 4 / 4

Discussion

35.	Find the valueo of a in the following: 6/3√2-2√3=3√2-a√3
Answer» Answer: ❍ Let Gaurav's and Sachin's ages one year ago be 6x and 7x respectively. Then, — Four years hence their ages; Gaurav's AGE = (6x + 1) + 4 = (6x + 5) years Sachin's age = (7x + 1) + 4 = (7x + 5) years \underline{\bigstar\:\boldsymbol{ACCORDING\; to \;the\; given\; Question :}} ★AccordingtothegivenQuestion: ⠀ According to given CONDITION, Four years hence the ratio of their ages (Gaurav's age & Sachin's age) would become 7: 8. ⠀ Therefore, ⠀ \begin{gathered}:\implies\sf \dfrac{(6x + 5)}{(7x + 5)} = \dfrac{7}{8} \\\\\\:\implies\sf 8(6x + 5) = 7(7x + 5) \\\\\\:\implies\sf 48x + 40 = 49x + 35\\\\\\:\implies\sf 48x - 49x = 35 - 40 \\\\\\:\implies\sf \cancel{\;-}\:1x = \cancel{\;-}\;5\\\\\\:\implies{\underline{\boxed{\frak{\pink{x = 5}}}}}\;\bigstar\end{gathered} :⟹ (7x+5) (6x+5) = 8 7 :⟹8(6x+5)=7(7x+5) :⟹48x+40=49x+35 :⟹48x−49x=35−40 :⟹ − 1x= − 5 :⟹ x=5 ★ ⠀ Hence, ⠀ Gaurav's present age = (6x + 1) = 6(5) + 1 = (30 + 1) = 31 years. Sachin's present age = (7x + 1) = 7(5) + 1 = (35 + 1) = 36 years. ⠀ \therefore{\underline{\textsf{Hence, \; Sachin\;is\;\textbf{36 years}\;OLD.}}}∴ Hence, Sachinis36 yearsold.

35.

Find the valueo of a in the following: 6/3√2-2√3=3√2-a√3

Answer»

Answer:

❍ Let Gaurav's and Sachin's ages one year ago be 6x and 7x respectively.

Then,

— Four years hence their ages;

Gaurav's AGE = (6x + 1) + 4 = (6x + 5) years

Sachin's age = (7x + 1) + 4 = (7x + 5) years

\underline{\bigstar\:\boldsymbol{ACCORDING\; to \;the\; given\; Question :}}

★AccordingtothegivenQuestion:

⠀

According to given CONDITION, Four years hence the ratio of their ages (Gaurav's age & Sachin's age) would become 7: 8.

⠀

Therefore,

⠀

\begin{gathered}:\implies\sf \dfrac{(6x + 5)}{(7x + 5)} = \dfrac{7}{8} \\\\\\:\implies\sf 8(6x + 5) = 7(7x + 5) \\\\\\:\implies\sf 48x + 40 = 49x + 35\\\\\\:\implies\sf 48x - 49x = 35 - 40 \\\\\\:\implies\sf \cancel{\;-}\:1x = \cancel{\;-}\;5\\\\\\:\implies{\underline{\boxed{\frak{\pink{x = 5}}}}}\;\bigstar\end{gathered}

:⟹

(7x+5)

(6x+5)

=

8

7

:⟹8(6x+5)=7(7x+5)

:⟹48x+40=49x+35

:⟹48x−49x=35−40

:⟹

−

1x=

−

5

:⟹

x=5

★

⠀

Hence,

⠀

Gaurav's present age = (6x + 1) = 6(5) + 1 = (30 + 1) = 31 years.

Sachin's present age = (7x + 1) = 7(5) + 1 = (35 + 1) = 36 years.

⠀

\therefore{\underline{\textsf{Hence, \; Sachin\;is\;\textbf{36 years}\;OLD.}}}∴

Hence, Sachinis36 yearsold.

Discussion

36.	Please friends give me answer. it's urgent.
Answer» ANSWER: OHHHHHH NAMASTE namaste namaste namaste namaste namaste

36.

Please friends give me answer. it's urgent.

Answer»

ANSWER:

OHHHHHH NAMASTE namaste namaste namaste namaste namaste

Discussion

37.	1. An amount of rupees 4500 is to be distributedamong Shrinath, Ramnath and Raghunath inthe ratio of 3: 7: 5 respectively. If 500 isadded to each of the share, the new ratio willbecome?A. 7: 10 : 13B. 9: 16 : 13C. 2:4:3D. 4 : 5:2
Answer» ★ How to do :- Here, we are given with the total MONEY that should be distributed with Shrinath, RAMNATH and Raghunath in the ratio of 3:7:5. We are asked to find the new ratio if some money i.e, ₹500 will be added to each of them, then the new ratio is to be found. So, first we should find the money they will get at first statement in 4500, then, we'll add 500 in each of their money to get the new ratio. So, let's solve!! $\:$ ➤ Solution :- ${\tt \leadsto 3:7:5 = 4500}$ ${\tt \leadsto 3x + 7x + 5x = 4500}$ ${\tt \leadsto 15x = 4500}$ ${\tt \leadsto x = \dfrac{4500}{15}}$ ${\tt \leadsto x = 300}$ $\:$ Now, multiply the VALUE of 'x' by each of the children's part of thier ratio. SHARES of Shrinath :- ${\tt \leadsto 3x = 3 \times 300}$ ${\tt \leadsto Rs \: \: 900}$ Shares of Ramnath :- ${\tt \leadsto 7x = 7 \times 300}$ ${\tt \leadsto Rs \: \: 2100}$ Shares of Raghunath :- ${\tt \leadsto 5x = 5 \times 300}$ ${\tt \leadsto Rs \: \: 1500}$ $\:$ Now, add 500 to each of their money to get the value of thier new ratio. New price of Shrinath :- ${\tt \leadsto 900 + 500}$ ${\tt \leadsto Rs \: \: 1400}$ New price of Ramnath :- ${\tt \leadsto 2100 + 500}$ ${\tt \leadsto Rs \: \: 2600}$ New price of Raghunath :- ${\tt \leadsto 1500 + 500}$ ${\tt \leadsto Rs \: \: 2000}$ $\:$ Now, WRITE all thier money in the form of ratio. ${\tt \leadsto 1400:2000:2600}$ ${\tt \leadsto 14 \cancel{00}:20 \cancel{00}:26 \cancel{00}}$ ${\tt \leadsto 14:20:26}$ ${\tt \leadsto \cancel{14}: \cancel{20}: \cancel{26}}$ ${\tt \leadsto \orange {\underline{\boxed{ \tt 7:10:13}}}}$ Hence solved !!

37.

1. An amount of rupees 4500 is to be distributedamong Shrinath, Ramnath and Raghunath inthe ratio of 3: 7: 5 respectively. If 500 isadded to each of the share, the new ratio willbecome?A. 7: 10 : 13B. 9: 16 : 13C. 2:4:3D. 4 : 5:2

Answer»

★ How to do :-

Here, we are given with the total MONEY that should be distributed with Shrinath, RAMNATH and Raghunath in the ratio of 3:7:5. We are asked to find the new ratio if some money i.e, ₹500 will be added to each of them, then the new ratio is to be found. So, first we should find the money they will get at first statement in 4500, then, we'll add 500 in each of their money to get the new ratio. So, let's solve!!

$\:$

➤ Solution :-

${\tt \leadsto 3:7:5 = 4500}$

${\tt \leadsto 3x + 7x + 5x = 4500}$

${\tt \leadsto 15x = 4500}$

${\tt \leadsto x = \dfrac{4500}{15}}$

${\tt \leadsto x = 300}$

$\:$

Now, multiply the VALUE of 'x' by each of the children's part of thier ratio.

SHARES of Shrinath :-

${\tt \leadsto 3x = 3 \times 300}$

${\tt \leadsto Rs \: \: 900}$

Shares of Ramnath :-

${\tt \leadsto 7x = 7 \times 300}$

${\tt \leadsto Rs \: \: 2100}$

Shares of Raghunath :-

${\tt \leadsto 5x = 5 \times 300}$

${\tt \leadsto Rs \: \: 1500}$

$\:$

Now, add 500 to each of their money to get the value of thier new ratio.

New price of Shrinath :-

${\tt \leadsto 900 + 500}$

${\tt \leadsto Rs \: \: 1400}$

New price of Ramnath :-

${\tt \leadsto 2100 + 500}$

${\tt \leadsto Rs \: \: 2600}$

New price of Raghunath :-

${\tt \leadsto 1500 + 500}$

${\tt \leadsto Rs \: \: 2000}$

$\:$

Now, WRITE all thier money in the form of ratio.

${\tt \leadsto 1400:2000:2600}$

${\tt \leadsto 14 \cancel{00}:20 \cancel{00}:26 \cancel{00}}$

${\tt \leadsto 14:20:26}$

${\tt \leadsto \cancel{14}: \cancel{20}: \cancel{26}}$

${\tt \leadsto \orange {\underline{\boxed{ \tt 7:10:13}}}}$

Hence solved !!

Discussion

38.	If surface area of a sphere is 676pa cm2, then its radius is
Answer» Step-by-step EXPLANATION: SURFACE AREA of SPHERE = 4 π r 2 4 π r 2 = 676 π r 2 = 676 4 = 169 r = 13 c m .

38.

If surface area of a sphere is 676pa cm2, then its radius is

Answer»

Step-by-step EXPLANATION:

SURFACE AREA of SPHERE

=

4

π

r

2

4

π

r

2

=

676

π

r

2

=

676

4

=

169

r

=

13

c

m

.

Discussion

39.	THE SCHOOL OPENS AT 7:30 AND CLOSES AT 2:15 PM. HOW LONG DOES THE SCHOOL FUNCATION.
Answer» Answer: The SCHOOL functions for 6 HOURS 45 minutes. Hope it helps you ✌️

39.

THE SCHOOL OPENS AT 7:30 AND CLOSES AT 2:15 PM. HOW LONG DOES THE SCHOOL FUNCATION.

Answer»

Answer:

The SCHOOL functions for 6 HOURS 45 minutes.

Hope it helps you ✌️

Discussion

40.	सही जोड़ी बनाइए।'अ''ब'1. पातालकोट।कूड़ादान2. अभिमन्युगीत3. भौंरायुद्ध 24. सेनाजनजाती अंचल5. कचराचक्रव्यूह
Answer» ANSWER: $Scientific notation is expressed in the form [latex]a \times 10^b[/latex] (where “b” is an integer and “a” is any real number), such as [latex]6.02 \times 10^{23}[/latex] .$

40.

सही जोड़ी बनाइए।'अ''ब'1. पातालकोट।कूड़ादान2. अभिमन्युगीत3. भौंरायुद्ध 24. सेनाजनजाती अंचल5. कचराचक्रव्यूह

Answer»

ANSWER:

$Scientific notation is expressed in the form [latex]a \times 10^b[/latex] (where “b” is an integer and “a” is any real number), such as [latex]6.02 \times 10^{23}[/latex] .$

Discussion

41.	Rs.4615 is divided into 3 parts and each part is lent at the rate of 15 percent per annum on simple interest for 4 6 and 8 years respectively.The interest received after the respective time periods on each part is equal.How much money is lent for 8 years on simple interest?
Answer» $\begin{gathered}\begin{gathered}\bf \:Let - \begin{cases} &\sf{first \: part \: be \: Rs \: x} \\ &\sf{second \: part \: be \: Rs \: y}\\ &\sf{third \: part \: be \: Rs \: z} \end{cases}\end{gathered}\end{gathered}$ So, $\bf\implies \:x \: + \: y \: + \: z \: = \: 4615 - - - (1)$ Now, $\begin{gathered}\begin{gathered}\bf \:Case \: 1 - \begin{cases} &\sf{Principal \: Rs \: x} \\ &\sf{Time, \: {t_{1}} = 4 \: years}\\ &\sf{<klux>RATE</klux>, \: {r_{1}} = 15\%} \end{cases}\end{gathered}\end{gathered}$ We know, $\: \: \: {\sf{\ Simple \: interest \: = \: \dfrac{P \times R \times T}{100}}}$ where, P represents Principal R represents Rate of interest T represents Time So, Simple interest in this case is $\rm :\longmapsto\:S.I._{1} \: = \: \dfrac{x \times 15 \times 4}{100} = \dfrac{60x}{100} - - - (2)$ $\begin{gathered}\begin{gathered}\bf \:Case \: 2 - \begin{cases} &\sf{Principal \: Rs \: y} \\ &\sf{Time, \: {t_{2}} = 6 \: years}\\ &\sf{Rate, \: {r_{2}} = 15\%} \end{cases}\end{gathered}\end{gathered}$ So, Simple interest in this case is $\rm :\longmapsto\:S.I._{2} \: = \: \dfrac{y \times 15 \times 6}{100} = \dfrac{90y}{100} - - - (3)$ $\begin{gathered}\begin{gathered}\bf \:Case \: 3 - \begin{cases} &\sf{Principal \: Rs \: z} \\ &\sf{Time, \: {t_{3}} = 8 \: years}\\ &\sf{Rate, \: {r_{3}} = 15\%} \end{cases}\end{gathered}\end{gathered}$ So, Simple interest in this case is $\rm :\longmapsto\:S.I._{3} \: = \: \dfrac{z \times 15 \times 8}{100} = \dfrac{120z}{100} - - - (4)$ According to statement, It is given that interest received after the RESPECTIVE time period on each part is same, $\bf\implies \:S.I._{1} = S.I._{2} = S.I._{3}$ $\rm :\implies\:\dfrac{60x}{100} = \dfrac{90y}{100} = \dfrac{120z}{100}$ $\: \: \: \: (\tt \: on \: dividing \: each \: term \: by \: \dfrac{30}{100} , \: we \: get)$ $\rm :\longmapsto\:2x = 3y = 4z$ $\sf \: Let \: \: 2x \: = \: 3y \: = \: 4z \: = \: k$ $\begin{gathered}\begin{gathered}\bf \:So, \: we \: have - \begin{cases} &\sf{x = \dfrac{k}{2} - - (5)} \\ &\sf{y = \dfrac{k}{3} - - (6) }\\ &\sf{z = \dfrac{k}{4} - - (7) } \end{cases}\end{gathered}\end{gathered}$ On substituting, the values of x, y, z in equation (1), we get $\rm :\longmapsto\:\dfrac{k}{2} + \dfrac{k}{3} + \dfrac{k}{4} = 4615$ $\rm :\longmapsto\:\dfrac{6k + 4k + 3k}{12} = 4615$ $\rm :\longmapsto\:\dfrac{13k}{12} = 4615$ $\bf\implies \:k = \dfrac{4615 \times 12}{13} = 4260 - - (8)$ $\begin{gathered}\begin{gathered}\bf \:So, \: value \: of \: each \: part - \begin{cases} &\sf{x = \dfrac{4260}{2} = Rs \: 2130} \\ &\sf{y = \dfrac{4260}{3} = Rs \: 1420 }\\ &\sf{z = \dfrac{4260}{4} = Rs \: 1065} \end{cases}\end{gathered}\end{gathered}$ So, it implies, Rs 1065 is lent for 8 years on Simple Interest. Additional Information :- Principal: The principal is the amount that initially borrowed from the bank or invested. The principal is denoted by P. Rate: Rate is the rate of interest at which the principal amount is given to SOMEONE for a certain time. The rate of interest is denoted by R. Time: Time is the duration for which the principal amount is given to someone. Time is denoted by T. Amount: When a person takes a loan from a bank, he/she has to return the principal borrowed plus the interest amount, and this total RETURNED is called Amount. Amount = Principal + Simple Interest

41.

Rs.4615 is divided into 3 parts and each part is lent at the rate of 15 percent per annum on simple interest for 4 6 and 8 years respectively.The interest received after the respective time periods on each part is equal.How much money is lent for 8 years on simple interest?

Answer»

$\begin{gathered}\begin{gathered}\bf \:Let - \begin{cases} &\sf{first \: part \: be \: Rs \: x} \\ &\sf{second \: part \: be \: Rs \: y}\\ &\sf{third \: part \: be \: Rs \: z} \end{cases}\end{gathered}\end{gathered}$

So,

$\bf\implies \:x \: + \: y \: + \: z \: = \: 4615 - - - (1)$

Now,

$\begin{gathered}\begin{gathered}\bf \:Case \: 1 - \begin{cases} &\sf{Principal \: Rs \: x} \\ &\sf{Time, \: {t_{1}} = 4 \: years}\\ &\sf{<klux>RATE</klux>, \: {r_{1}} = 15\%} \end{cases}\end{gathered}\end{gathered}$

We know,

$\: \: \: {\sf{\ Simple \: interest \: = \: \dfrac{P \times R \times T}{100}}}$

where,

P represents Principal

R represents Rate of interest

T represents Time

So,

Simple interest in this case is

$\rm :\longmapsto\:S.I._{1} \: = \: \dfrac{x \times 15 \times 4}{100} = \dfrac{60x}{100} - - - (2)$

$\begin{gathered}\begin{gathered}\bf \:Case \: 2 - \begin{cases} &\sf{Principal \: Rs \: y} \\ &\sf{Time, \: {t_{2}} = 6 \: years}\\ &\sf{Rate, \: {r_{2}} = 15\%} \end{cases}\end{gathered}\end{gathered}$

So, Simple interest in this case is

$\rm :\longmapsto\:S.I._{2} \: = \: \dfrac{y \times 15 \times 6}{100} = \dfrac{90y}{100} - - - (3)$

$\begin{gathered}\begin{gathered}\bf \:Case \: 3 - \begin{cases} &\sf{Principal \: Rs \: z} \\ &\sf{Time, \: {t_{3}} = 8 \: years}\\ &\sf{Rate, \: {r_{3}} = 15\%} \end{cases}\end{gathered}\end{gathered}$

So, Simple interest in this case is

$\rm :\longmapsto\:S.I._{3} \: = \: \dfrac{z \times 15 \times 8}{100} = \dfrac{120z}{100} - - - (4)$

According to statement,

It is given that interest received after the RESPECTIVE time period on each part is same,

$\bf\implies \:S.I._{1} = S.I._{2} = S.I._{3}$

$\rm :\implies\:\dfrac{60x}{100} = \dfrac{90y}{100} = \dfrac{120z}{100}$

$\: \: \: \: (\tt \: on \: dividing \: each \: term \: by \: \dfrac{30}{100} , \: we \: get)$

$\rm :\longmapsto\:2x = 3y = 4z$

$\sf \: Let \: \: 2x \: = \: 3y \: = \: 4z \: = \: k$

$\begin{gathered}\begin{gathered}\bf \:So, \: we \: have - \begin{cases} &\sf{x = \dfrac{k}{2} - - (5)} \\ &\sf{y = \dfrac{k}{3} - - (6) }\\ &\sf{z = \dfrac{k}{4} - - (7) } \end{cases}\end{gathered}\end{gathered}$

On substituting, the values of x, y, z in equation (1), we get

$\rm :\longmapsto\:\dfrac{k}{2} + \dfrac{k}{3} + \dfrac{k}{4} = 4615$

$\rm :\longmapsto\:\dfrac{6k + 4k + 3k}{12} = 4615$

$\rm :\longmapsto\:\dfrac{13k}{12} = 4615$

$\bf\implies \:k = \dfrac{4615 \times 12}{13} = 4260 - - (8)$

$\begin{gathered}\begin{gathered}\bf \:So, \: value \: of \: each \: part - \begin{cases} &\sf{x = \dfrac{4260}{2} = Rs \: 2130} \\ &\sf{y = \dfrac{4260}{3} = Rs \: 1420 }\\ &\sf{z = \dfrac{4260}{4} = Rs \: 1065} \end{cases}\end{gathered}\end{gathered}$

So, it implies,

Rs 1065 is lent for 8 years on Simple Interest.

Additional Information :-

Principal: The principal is the amount that initially borrowed from the bank or invested. The principal is denoted by P.

Rate: Rate is the rate of interest at which the principal amount is given to SOMEONE for a certain time. The rate of interest is denoted by R.

Time: Time is the duration for which the principal amount is given to someone. Time is denoted by T.

Amount: When a person takes a loan from a bank, he/she has to return the principal borrowed plus the interest amount, and this total RETURNED is called Amount.

Amount = Principal + Simple Interest

Discussion

42.	A well whose diemeter is 14 m and 25 m deep.findthe volume of mud taken out of the well.Also find cost of plastering the well from insiderate of Rs. 3 per squaremeter.
Answer» Answer: HERES your answer! Step-by-step explanation: volume=πr²h =(22/7 X 7 x 7 x 2500)cm³ =385000cm³ cost of plastering=2πrh×3 =2 x 22/7 x 7/100 x 25 x 3 =33Rupees please mark as BRAINLIES

42.

A well whose diemeter is 14 m and 25 m deep.findthe volume of mud taken out of the well.Also find cost of plastering the well from insiderate of Rs. 3 per squaremeter.

Answer»

Answer:

HERES your answer!

Step-by-step explanation:

volume=πr²h

=(22/7 X 7 x 7 x 2500)cm³

=385000cm³

cost of plastering=2πrh×3

=2 x 22/7 x 7/100 x 25 x 3

=33Rupees

please mark as BRAINLIES

Discussion

43.	Write notice event organized in school
Answer» Answer: ⚘Answer⚘⤵️⤵️ Step-by-step explanation: NAME and place of the school, organisation or OFFICE issuing the notice should be MENTIONED. Give an appropriate heading. Write the DATE of issuing the notice. ... Signature, name and designation of the person issuing the notice. ☺️☺️

43.

Write notice event organized in school

Answer»

Answer:

⚘Answer⚘⤵️⤵️

Step-by-step explanation:

NAME and place of the school, organisation or OFFICE issuing the notice should be MENTIONED. Give an appropriate heading. Write the DATE of issuing the notice. ... Signature, name and designation of the person issuing the notice.

☺️☺️

Discussion

44.	Today's ThroughThe Struggle Your in today is developing the Stregth you need for Tomorrow. If you wish to be a friend of me come on idj-quuk-dwt
Answer» Step-by-step EXPLANATION: ayamyalyrayl5sny ED U TX duttsutsltskyskdilso6sd7

44.

Today's ThroughThe Struggle Your in today is developing the Stregth you need for Tomorrow. If you wish to be a friend of me come on idj-quuk-dwt

Answer»

Step-by-step EXPLANATION:

ayamyalyrayl5sny ED U TX duttsutsltskyskdilso6sd7

Discussion

45.	Diameter of a circle =2Xr[T/F]
Answer» ANSWER: TRUE is the CORRECT answer for this QUESTION.

45.

Diameter of a circle =2Xr[T/F]

Answer»

ANSWER:

TRUE is the CORRECT answer for this QUESTION.

Discussion

46.	Please I need help with this
Answer» ANSWER: RIGTH answer is Step-by-step explanation: 1.p 2.r 3.i 4.n 5.c 6.e prince

46.

Please I need help with this

Answer»

ANSWER:

RIGTH answer is

Step-by-step explanation:

1.p

2.r

3.i

4.n

5.c

6.e

prince

Discussion

47.	How many days are there from May 7 to August 23?
Answer» ANSWER: HERES your answer hope it helps you! Step-by-step EXPLANATION: there are 108 days from may 7 to august 23 please MARK the BRAINLIEST

47.

How many days are there from May 7 to August 23?

Answer»

ANSWER:

HERES your answer hope it helps you!

Step-by-step EXPLANATION:

there are 108 days from may 7 to august 23

please MARK the BRAINLIEST

Discussion

48.	B) Write short note on any two sources to know history
Answer» Answer: Historical sources include DOCUMENTS, ARTIFACTS, archaeological sites, features. oral transmissions, stone inscriptions, paintings, recorded sounds, IMAGES (photographs, MOTION PICTURE), and oral history. Even ancient relics and ruins, broadly speaking, are historical sources.

48.

B) Write short note on any two sources to know history

Answer»

Answer:

Historical sources include DOCUMENTS, ARTIFACTS, archaeological sites, features. oral transmissions, stone inscriptions, paintings, recorded sounds, IMAGES (photographs, MOTION PICTURE), and oral history. Even ancient relics and ruins, broadly speaking, are historical sources.

Discussion

49.	Find ratio in simplest form 25p to ₹ 6.25
Answer» Step-by-step explanation: 1 rupee = 100 paisa 6.25 rupee = 625 paisa 25:625 = 1:25 hope it HELPS you ! Mark me as the BRAINLIEST !

49.

Find ratio in simplest form 25p to ₹ 6.25

Answer»

Step-by-step explanation:

1 rupee = 100 paisa

6.25 rupee = 625 paisa

25:625 = 1:25

hope it HELPS you !

Mark me as the BRAINLIEST !

Discussion

50.	17+2x >11, where x is a negative integer.
Answer» ANSWER: I HOPE this answer will HELP you

50.

17+2x >11, where x is a negative integer.

Answer»

ANSWER:

I HOPE this answer will HELP you

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

out of men and 3women ,a committe of three members is to be formed so that it has 1 women ,and 2men .in how many diffrent ways can it be done​

Higest marks-lowest marks​

Compare √5 upon 3 ,3 upon √7​

The factors of 36x2 – 25 are

+++Convert ento decimal fracking 500g​

In the given figures, measure of some parts of the triangle are indicated.By applying SAS congruence rule.State the pair of congruent triangles, if any.In case of congruent triangles, write them in symbolic form.​

A boy runs 1 km in 10 minutes. How long will he take to ran 600m?​

Find the difference between 31m and 8 m 86 cm​

Subtract 10xy from 14xy.​

Find the quadratic polynomial whose one zero is 3and product of zero is 12​

Find the area of rectangle whose length is 10 cm and breath 50 cm​

✭ CORRECT Question :

✭ Required SOLUTION :

✦ FORMULA used here :

✭ Value given to us :

✭ Putting values in formula :

✰ Area of Rectangle is 500 cm² ✰

___________________________

55-11 [120:24 +10+5-7)​

Ram bought a second hand cycle for 500. He spent 80 in repairing and175 in repainting. He then sold it to Lalit for 900. How much did he gain orlose?​

Differentiate the function w. r. t x e^x / 1+x^2​

Let y=

(X−3)(x−4)(x−5)

(x−1)(x−2)

TAKING logarithm on both the sides, we obtain

logy=log

(x−3)(x−4)(x−5)

(x−1)(x−2)

⇒logy=

2

1

log[

(x−3)(x−4)(x−5)

(x−1)(x−2)

]

⇒logy=

2

1

[log[(x−1)(x−2)]−log[(x−3)(x−4)(x−5)]]

⇒logy=

2

1

[log(x−1)+log(x−2)−log(x−3)−log(x−4)−log(x−5)]

Differentiating both sides with respect to x, we obtain

⇒

y

1

dx

dy

=

2

1

[

x−1

1

.+

x−2

1

−

x−3

1

−

x−4

1

−

x−5

1

]

⇒

dx

dy

=

2

y

(

x−1

1

+

x−2

out of men and 3women ,a committe of three members is to be formed so that it has 1 women ,and 2men .in how many diffrent ways can it be done

Higest marks-lowest marks

Compare √5 upon 3 ,3 upon √7

+++Convert ento decimal fracking 500g

In the given figures, measure of some parts of the triangle are indicated.By applying SAS congruence rule.State the pair of congruent triangles, if any.In case of congruent triangles, write them in symbolic form.

A boy runs 1 km in 10 minutes. How long will he take to ran 600m?

Find the difference between 31m and 8 m 86 cm

Subtract 10xy from 14xy.

Find the quadratic polynomial whose one zero is 3and product of zero is 12

Find the area of rectangle whose length is 10 cm and breath 50 cm

55-11 [120:24 +10+5-7)

Ram bought a second hand cycle for 500. He spent 80 in repairing and175 in repainting. He then sold it to Lalit for 900. How much did he gain orlose?

Differentiate the function w. r. t x e^x / 1+x^2