332354 + Interview Questions in Secondary School in Math

1.	use division algorithm to show that the cube of any positive integer is of the form 9m or 9m+1 or 9m+8
Answer» We know, For every POSITIVE integer a and b there always exists another two integers q and r such that a=bq+r and 0<=r Taking b=3 a=3q+0 or a=3q+1 or a=3q+2 Now, a^3=(3q)^3 =27q^3 =9×3q^3 =9m[Taking 3q^3=m] a^3=(3q+1)^3 =(3q)^3 + 3×(3q)^2×1 + 3×3q×1^2+1^3 =27q^3 + 27q^2 +9q+1 =9(3q^3+3q^2+q)+1 =9m+1[Taking 3q^3+3q^2+q=m] a^3=(3q+2)^3 =(3q)^3 + 3×(3q)^2×2 + 3×3q×2^2+2^3 =27q^3 + 54q^2 +36q+8 =9(3q^3+6q^2+4q)+8 =9m+8[Taking 3q^3+6q^2+4q=m] proved.... hope it helps u....

Discussion

2.	93x=12-15y,Express the given linear equation in the form of ax+by+c=0 and indicate the values of a, b and c in each case.
Answer» $\bold{Answer :}$ Given, 93X = 12 - 15y or, 93x + 15y - 12 = 0 or, 93x + 15y + (- 12) = 0 ...(i) Now, COMPARING (i) with ax + by + c = 0, we get a = 93, b = 15 and c = - 12 # $\bold{MarkAsBrainliest}$

Discussion

3.	Given √3=1.732app.,find to three places of decimal the value of 1+2√3/2-√3
Answer» 1+2root 3/2-root 3=1 +2×1.732/2-1.732 =1+3.464/0.268=4.464/0.268=4464×1000/268×1000=4464/268 (100/100 is CANCELLED) =16.909 APPROX.

Discussion

4.	List five axioms (postulates) used in this book.
Answer» The five postulates are :- A straight line may be drawn from any one point to any other point. (2) A TERMINATED line can be produced indefinitely . (3) A circle can be drawn with any center and any radius.(4) All right angles are EQUAL to one another. (5) If straight line falling on two straight LINES makes the INTERIOR angles on the same sides of it taken together less than two right angles, then the two straight lines if produced indefinitely meet on that side which the sum of angles is less than two right angles....

Discussion

5.	The area of square field is 37056.25.sq .m.Find the square field
Answer» HEYA !! AREA of SQUARE = 37056.25 ( SIDE)² = 37056.25 Side = ✓37056.25 Side = 192.5 m. Hence, Side of square FIELD = 192.5 m.

Discussion

6.	Find the square root of 7+√48
Answer» *************************************************** Hope u UNDERSTAND my words i NEED only few BRAINLIEST for the next RANK. If i was helpful then pl pl pl MARK me as brainliest

Discussion

7.	Factories these and give answer (14m-21).
Answer» The ANSWER is 7(2m-3). HOPE THIS HELPS THANK YOU :))

Discussion

8.	If (1+i)z =(1-i)z conjugated show z =-iz conjugated
Answer» ( 1 + i)Z = (1 - i)z' Now, To PROVE , z= -z' Proof: So, 1 + i in POLAR form can be written as, √2 (e^(i×pi) / 2) And, 1- i = √2(e^(-i×pi) / 2) So, ( 1 +i)z = (1 - i)z' √2 (e^(i×pi) / 2) z = √2(e^(-i×pi) / 2)z' z= (e^(-i×pi/2 - i× pi/ 2))z' z= (e^(-i×pi)) z' Now, (e^(-i×pi) = cos (pi) + ISIN (pi) = -1 So, z= -z'

8.

If (1+i)z =(1-i)z conjugated show z =-iz conjugated

Answer»

( 1 + i)Z = (1 - i)z'

Now,

To PROVE ,

z= -z'

Proof:

So,

1 + i in POLAR form can be written as,

√2 (e^(i×pi) / 2)

And,

1- i = √2(e^(-i×pi) / 2)

So,

( 1 +i)z = (1 - i)z'

√2 (e^(i×pi) / 2) z = √2(e^(-i×pi) / 2)z'

z= (e^(-i×pi/2 - i× pi/ 2))z'

z= (e^(-i×pi)) z'

Now,

(e^(-i×pi) = cos (pi) + ISIN (pi)

= -1

So,

z= -z'

Discussion

9.	The sum of 2nd and 7th term of AP is 30 if its 15 term is 1 less than twice its a term find the AP
Answer» Here is your SOLUTION GIVEN:- t2+t7=30 =>a+d+a+6d=30 =>2a+7d=30..................... (1) A/q 15th term=2×8th term-1 =>a+14d=2×(a+7d)-1 =>a+14d=2a+14d-1 =>14d-14d+1=2a-a =>1=a Putting the value of a in equation (1) 2a×7d=30 2×1+7d=30 2+7d=30 7d=30-2 7d=28 d=4 A.p are a,a+d,a+2d,a+3d,a+4d =>A.p are 1,5,9,13......... HOPE IT HELPS YOU

Discussion

10.	if one zero of the quadratic polynomial 2x^2 - 3x -p is 3, find its other zero. Also find the value of p
Answer» P(3)=2(3)^2-3(3)-p=0 18-9-p=0 9-p=0 9=p p=9 other ZEROES, => 2X^2-6x+3x-9=0 2x(x-3)+3(x-3)=0 (2x+3)(x-3)=0 other zero, 2x+3=0 2x=-3 x=(-3)/2

Discussion

11.	एक cricket team k 12 members ki average age 30 year hai unke sath captain ki shamil ke lene pr yah average age 1 year. badh jatei hai captain ki age batayi
Answer» CAPTAIN AGE 31 HOGI

Discussion

12.	What is the angle between the two hand of a clock (approximately) when the time is 9:30?
Answer» 90 degree I am sure I READ in CLASS FIVE

Discussion

13.	Can someone tell me how to get my math sums correct
Answer» These are the few steps to create WONDERS in maths ---------------------------------------------------- (1) Learn all the formulas by heart.. (2)WRITE down the formulas in ur practice NOTEBOOK. (3)Do practice 20 SUMS daily. (4)After completing the sums do revise them atleast once. (5)It would be more better if u do by time.. Hope this HELPED.. plz Mark as brainliest

Discussion

14.	X²+5x+6, Factorise it using appropriate identities.
Answer» SOLUTION : Given Quadratic EXPRESSION is x² + 5x + 6 splitting the middle term , we GET = x² + 2x + 3x + 6 = X( x + 2 ) + 3( x + 2 ) = ( x + 2 )( x + 3 ) •••••

Discussion

15.	By what rational number should -8/39 be multiplied to obtain 5/26
Answer» The no. by which -8/39 be MULTIPLIED to OBTAIN 5/26 is :-(5/26)/(-8/39) =(5/26)*(-39/8) =-195/208

Discussion

16.	(-2a+5b-3c)²,Expand it, using suitable identities.
Answer» SOLUTION : *************************************** We KNOW the ALGEBRAIC IDENTITY, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx *************************************** Now , ( -2a + 5b - 3c )² = [ (-2a) + 5b + ( -3c ) ]² = (-2a)²+(5b)²+(-3c)²+2(-2a)(5b) +2(5b)(-3c)+2(-3c)(-2a) = 4a²+25b²+9c²-20ab-30bc+12ca •••••••

Discussion

17.	Price of an article is first increased by 40% and then reduced by 20%. What is the percentage increase in its price
Answer» Let the original PRICE be Rs. 100. New FINAL price = 120 % of (75 % of Rs. 100) = Rs. (120/100 * 75/100 * 100) = Rs. 90. Decrease = 10% Do ur SUM in this method

Discussion

18.	A fruit seller buys oranges at 4 for rd 3 and sells them at 3 for 4 .find his profit persent
Answer» His profit is one RUPEES because he BUY 4FOR 3 and sell 3FOR 4

Discussion

19.	Find the rationalising factor of root 5 minus root 3
Answer» ROOT 5+root3 is the RATIONALISING FACTOR

Discussion

20.	Find the volume of the cylinder if height (h) and radius of the base (r) are as given below.(1) r =10.5 cm, h = 8 cm (2) r = 2.5 m, h = 7 m(3) r = 4.2 cm, h = 5 cm (4) r = 5.6 cm, h = 5 cm
Answer» Answer: Step-by-step EXPLANATION: Volume of cylinder = π r² h R = radius h = height π = 22/7 (1) r =10.5 cm, h = 8 cm Volume = (22/7)(10.5)²* 8 = 2772 cm³ (2) r =2.5 cm, h = 7 cm Volume = (22/7)(2.5)²* 7 = 137.5 cm³ (3) r =4.2 cm, h = 5 cm Volume = (22/7)(4.2)²* 5 = 277.2 cm³ (4) r =5.6 cm, h = 5 cm Volume = (22/7)(5.6)²* 5 = 492.8 cm³

20.

Find the volume of the cylinder if height (h) and radius of the base (r) are as given below.(1) r =10.5 cm, h = 8 cm (2) r = 2.5 m, h = 7 m(3) r = 4.2 cm, h = 5 cm (4) r = 5.6 cm, h = 5 cm

Answer»

Answer:

Step-by-step EXPLANATION:

Volume of cylinder = π r² h

R = radius

h = height

π = 22/7

(1) r =10.5 cm, h = 8 cm

Volume = (22/7)(10.5)²* 8 = 2772 cm³

(2) r =2.5 cm, h = 7 cm

Volume = (22/7)(2.5)²* 7 = 137.5 cm³

(3) r =4.2 cm, h = 5 cm

Volume = (22/7)(4.2)²* 5 = 277.2 cm³

(4) r =5.6 cm, h = 5 cm

Volume = (22/7)(5.6)²* 5 = 492.8 cm³

Discussion

21.	What fraction do the blue segments make up?
Answer» 4/8 + 3/8 7/8 ANS or 4+3 / 8+8 7/16 ANS

Discussion

22.	if Q is the midpoint of a line segment a b and P is the midpoint of a q then show that p q is equal to one by fourth of a b
Answer» <P>AQ =QB (q is the MIDPOINT of AB ) AQ +QB = AB 2AQ = AB AQ =1/2 AB AP =PQ ( P is tge midpoint of AQ ) AP + PQ = AQ 2AP = AQ AP = 1/2 AQ As proved EARLIER that AQ =1/2AB So replacing VALUE o AQ AP = 1/2 (1/2 AB ) AP =1/4AB

Discussion

23.	If a+b+c=15, a²+b²+c²=83. Then find a³+b³+c³-3abc
Answer» A^2 + b^2 + c^2 = 83 ------(1) a + b+ c = 15 On squaring both sides, we get a^2 + b^2 + c^2 + 2( ab + bc + CA) = 225 => 83 + 2 ( ab + bc + ca) = 225 => 2 ( ab + bc + ca) = 142 => ab + bc + ca = 71 --------(2) Now, a^3 + b^3 + c^3 - 3abc = ( a +b + c) (a^2 + b^2 + c^2 - ab - bc - ca) = ( 15) [ 83 - (ab + bc + ca)] = (15)(83-71) = 15 × 12 = 180

Discussion

24.	The angels of quadrilateral are in ratio 3:5 : 9:13
Answer» LET an ANGLE be x 3x + 5X + 9x + 13x = 30x 30x = 360° x = 360°/ 30 x = 12° 3x = 12°×3 = 36° 5x = 12°×5 = 60° 9x = 12°×9 = 108° 13x = 12°×13 = 156°

Discussion

25.	If a^2 + 1/a^2 =18 then find : a-1/a
Answer» ANSWER: $a-\frac{1}{a}=\pm 4$ Step-by-step explanation: Given : $a^2+\frac{1}{a^2}=18$ To FIND : The value of $a-\frac{1}{a}$ ? Solution : Using IDENTITY, $(x-y)^2=x^2+y^2-2xy$ LET $x=a$ and $y=\frac{1}{a}$ Substitute in the identity, $(a-\frac{1}{a})^2=a^2+\frac{1}{a^2}-2\times a\times \frac{1}{a}$ $(a-\frac{1}{a})^2=18-2$ $(a-\frac{1}{a})^2=16$ TAKING root both side, $a-\frac{1}{a}=\pm 4$

25.

If a^2 + 1/a^2 =18 then find : a-1/a

Answer»

ANSWER:

$a-\frac{1}{a}=\pm 4$

Step-by-step explanation:

Given : $a^2+\frac{1}{a^2}=18$

To FIND : The value of $a-\frac{1}{a}$ ?

Solution :

Using IDENTITY,

$(x-y)^2=x^2+y^2-2xy$

LET $x=a$ and $y=\frac{1}{a}$

Substitute in the identity,

$(a-\frac{1}{a})^2=a^2+\frac{1}{a^2}-2\times a\times \frac{1}{a}$

$(a-\frac{1}{a})^2=18-2$

$(a-\frac{1}{a})^2=16$

TAKING root both side,

$a-\frac{1}{a}=\pm 4$

Discussion

26.	P is a point in the interior of a parallelogram ABCD. Show that(i) ar(ΔAPB)+ar(ΔPCD)=1/2ar(ABCD)(ii) ar(ΔAPD)+ar(ΔPBC)=ar(ΔAPB)+ar(ΔPCD)(Hint : Through P, draw a line parallel to AB)
Answer» If a parallelogram and a triangle are on the same base and between the same parallels then area of the triangle is half the area of the parallelogram. ================================================================ Given: ABCD is a parallelogram So, AB\|\|CD & AD\|\| BC To show: (i) ar (APB) + ar (PCD) = ar (ABCD) (ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD) Proof: (i)Through the point P ,draw GH parallel to AB. In a parallelogram, AB \|\| GH (by construction) — (i) Thus, AD \|\| BC ⇒ AG \|\| BH — (ii) From equations (i) and (ii), ABHG is a parallelogram. Now, In ΔAPB and parallelogram ABHG are lying on the same base AB and between the same parallel lines AB and GH. ∴ ar(ΔAPB) = 1/2 ar(ABHG) — (iii) also, In ΔPCD and parallelogram CDGH are lying on the same base CD and between the same parallel lines CD and GH. ∴ ar(ΔPCD) = 1/2 ar(CDGH) — (IV) Adding equations (iii) and (iv), ar(ΔAPB) + ar(ΔPCD) = 1/2 {ar(ABHG) + ar(CDGH)} ar(APB) + ar(PCD) = 1/2 ar(ABCD) (ii) A line EF is drawn parallel to AD passing through P. In a parallelogram, AD \|\| EF (by construction) — (i) Thus, AB \|\| CD ⇒ AE \|\| DF — (ii) From equations (i) and (ii), AEDF is a parallelogram. Now, In ΔAPD and parallelogram AEFD are lying on the same base AD and between the same parallel lines AD and EF. ∴ ar(ΔAPD) = 1/2 ar(AEFD) — (iii) also, In ΔPBC and parallelogram BCFE are lying on the same base BC and between the same parallel lines BC and EF. ∴ ar(ΔPBC) = 1/2 ar(BCFE) — (iv) Adding equations (iii) and (iv), ar(ΔAPD) + ar(ΔPBC) = 1/2 {ar(AEFD) + ar(BCFE)} ar(ΔAPD) + ar(ΔPBC) =1/2 ar(ABCD) ar(APD) + ar(PBC) = ar(APB) + ar(PCD) ( From PART i) Hope this will help you...

26.

P is a point in the interior of a parallelogram ABCD. Show that(i) ar(ΔAPB)+ar(ΔPCD)=1/2ar(ABCD)(ii) ar(ΔAPD)+ar(ΔPBC)=ar(ΔAPB)+ar(ΔPCD)(Hint : Through P, draw a line parallel to AB)

Answer»

If a parallelogram and a triangle are on the same base and between the same parallels then area of the triangle is half the area of the parallelogram.

================================================================

Given: ABCD is a parallelogram

So, AB||CD & AD|| BC

To show:

(i) ar (APB) + ar (PCD) = ar (ABCD)

(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

Proof:

(i)Through the point P ,draw GH parallel to AB.

In a parallelogram,

AB || GH (by construction) — (i)

Thus,

AD || BC ⇒ AG || BH — (ii)

From equations (i) and (ii),

ABHG is a parallelogram.

Now,

In ΔAPB and parallelogram ABHG are lying on the same base AB and between the same parallel lines AB and GH.

∴ ar(ΔAPB) = 1/2 ar(ABHG) — (iii)

also,

In ΔPCD and parallelogram CDGH are lying on the same base CD and between the same parallel lines CD and GH.

∴ ar(ΔPCD) = 1/2 ar(CDGH) — (IV)

Adding equations (iii) and (iv),

ar(ΔAPB) + ar(ΔPCD) = 1/2 {ar(ABHG) + ar(CDGH)}

ar(APB) + ar(PCD) = 1/2 ar(ABCD)

(ii)

A line EF is drawn parallel to AD passing through P.

In a parallelogram,

AD || EF (by construction) — (i)

Thus,

AB || CD ⇒ AE || DF — (ii)

From equations (i) and (ii),

AEDF is a parallelogram.

Now,

In ΔAPD and parallelogram AEFD are lying on the same base AD and between the same parallel lines AD and EF.

∴ ar(ΔAPD) = 1/2 ar(AEFD) — (iii)

also,

In ΔPBC and parallelogram BCFE are lying on the same base BC and between the same parallel lines BC and EF.

∴ ar(ΔPBC) = 1/2 ar(BCFE) — (iv)

Adding equations (iii) and (iv),

ar(ΔAPD) + ar(ΔPBC) = 1/2 {ar(AEFD) + ar(BCFE)}

ar(ΔAPD) + ar(ΔPBC) =1/2 ar(ABCD)

ar(APD) + ar(PBC) = ar(APB) + ar(PCD) ( From PART i)

Hope this will help you...

Discussion

27.	The mean (average) weight of three students is 40 kg. One of the students Ranga weighs 46 kg. The other two students, Rahim and Reshma have the same weight. Find Rahims weight.
Answer» LET WEIGHT of RAHIM =weight of reshma=X hence from GIVEN data 40+x+x/3=46 40+2x=46×3 2x=138-40 2x=98 hence x= rahim's weight=98/2=49

Discussion

28.	One fourth of a number is 5 more than 6 find the number
Answer» LET the NUMBER be X $x \div 4 = 5 + 6 \\ x = 11 \times 4 \\ x = 44$

Discussion

29.	if the school bag of neelam and garima weight 6 kg 265 g and 5kg 80 g whose bag is heavier and how much
Answer» Obviously neelam bag is HEAVIER than GARIMA bag by (1kg=1000gm)(6KG 265gm-5kg 80 gm=6265-5080=1185 gm) 1kg 185 gm .

Discussion

30.	by selling 80 ball pens for 80 rupees a retailer loses 30%. how many ball pens should he sell for 104 rupees so as to make a profit of 30%
Answer» 32 Description for CORRECT answer According to question, \\ CP of 80 BALLS pens = Rs. 200 CP of 1 ball pens = 2008020080 = Rs. 2.5 to gain 30% SP of 1 ball PEN = 2.5+30100×3.252.5+30100×3.25 Ball pens he sale in Rs. 104 = 3.25 = 32 Pens

Discussion

31.	In the given figure if AB\|\|DE, ∠BAC=35° and ∠CDE=53°, find ∠DCE.
Answer» In the figure, because AB\|\|DE So, ANGLES) Now, in the ∆DCE, the total add of interior angles is 180°. So, Hope it was helpful. Please MARK it as BRAINLIEST.

Discussion

32.	A cricketer has a certain average for 10 innings. in the eleventh inning, he scored 108runs, therebyincreasing his average by 6 rund. his new average is
Answer» Let X be the average So runs scored in first 10 innings = 10X Runs scored in 11th inning = 108 Average after 11th inning is increased by 6 so average after 11th inning is x+6 Runs after 11th inning is 11(x+6) as PER the question, 10x + 108 = 11 (x+6) 10x + 108 = 11x + 66 x = 42 his old average is 42 new average is 42 + 6 = 48

32.

A cricketer has a certain average for 10 innings. in the eleventh inning, he scored 108runs, therebyincreasing his average by 6 rund. his new average is

Answer»

Let X be the average

So runs scored in first 10 innings = 10X

Runs scored in 11th inning = 108

Average after 11th inning is increased by 6 so average after 11th inning is x+6

Runs after 11th inning is 11(x+6)

as PER the question,

10x + 108 = 11 (x+6)

10x + 108 = 11x + 66

x = 42

his old average is 42

new average is 42 + 6 = 48

Discussion

33.	The ratio of earning of x and y 13:15 .if the earning of x increase by 20% and the earning of y decrease
Answer» REDUCE by 4% . .....is the ANSWER..

Discussion

34.	Without actually calculating find the value of 25 cube minus 75 cube + 50 cube
Answer» $take \: {25}^{3} common \: from \: each \: term$ $then \: we \: get \: {25}^{3} (1 - {3}^{3} + {2}^{3})$ $now \: we \: can \: easily \: simpify \:$ ${25}^{3} (1 - 27 + 8) = {25}^{3}( - 18)$ ${25}^{3} = 15625 \:$ $answer \: is \: 15625( - 18) \: = - 281250$ $i \: tried \: m y \: level \: best \: \: hope$ $it \: helps \: you$

Discussion

35.	2nc3: nc3=11:1 the find value of n
Answer» Here is your ANSWER. N is EQUAL to 6

Discussion

36.	If the number 653xy is divisible by 90 then x+y=?
Answer» If the number 653xy is divisible by 90 so, 653xy should be divisible by both 9 and 10 any number divisible by 10 should have ZERO at END of it's number so, y = 0 so,the number is 653x0 any number divisible by 9 sum of it's digits should be divisible by 9 so, 6+5+3+0 = 14 divisible by 9 which will remainder of 5 so, 5+4 = 9 which divisible by 9 so, x = 4 so, the number is 65340 which is divisible by 90 x + y = 4+0 = 4

Discussion

37.	a train 150m long takes 30sec to cross a dridge 500m long. how much time will the train take to cross a platfor 370m long
Answer» HEYA MATE, HERE IS UR ANSWER. Thanks for asking this question. When a TRAIN CROSES a bridge, distance covered = length of (bridge + train). therefore, SPEED of train= 150+500/30 = 650/30 =65/3 m/sec Therefore, Time taken to cross the 370m long platform = 370+150/65/3 =520×3/65 =18sec I HOPE IT HELPS U.

Discussion

38.	CosA -cos(90-A)what is the value of above term
Answer» COS A -cos 90-A=cosA-cotA =0

Discussion

39.	Find the area of a triangle two sides of which are 18cm and 10 cm and the perimeter is 42 cm
Answer» The AREA of this TRIANGLE is 180CM

Discussion

40.	The value of this is sin(2sin^-1(0.6))
Answer» Answer: 0.96 Step-by-step EXPLANATION: GIVEN, $sin(2sin^{-1}(0.6))$ LET, $sin^{-1}(0.6)=A$ -------(1) $\implies sin A = 0.6$ $\implies \sqrt{1-cos^2A} = 0.6$ ( SINCE, sin² A + cos² A = 1 ⇒ sin² A = 1 - cos² A ⇒ sin A = √(1-cos² A) ) $1-cos^2A = 0.36$ $cos^2A= 1-0.36 = 0.64$ $\implies cos A = 0.8$ We know that, $sin 2A = 2 sin A cos A = 2\times 0.6\times 0.8=0.96$ From equation (1), $\implies sin 2[sin^{-1}(0.6)]=0.96$

40.

The value of this is sin(2sin^-1(0.6))

Answer»

Answer:

0.96

Step-by-step EXPLANATION:

GIVEN,

$sin(2sin^{-1}(0.6))$

LET,

$sin^{-1}(0.6)=A$ -------(1)

$\implies sin A = 0.6$

$\implies \sqrt{1-cos^2A} = 0.6$ ( SINCE, sin² A + cos² A = 1 ⇒ sin² A = 1 - cos² A ⇒ sin A = √(1-cos² A) )

$1-cos^2A = 0.36$

$cos^2A= 1-0.36 = 0.64$

$\implies cos A = 0.8$

We know that,

$sin 2A = 2 sin A cos A = 2\times 0.6\times 0.8=0.96$

From equation (1),

$\implies sin 2[sin^{-1}(0.6)]=0.96$

Discussion

41.	Algebraic identities evaluate X if: 6.2x=8.1×8.1-1.9×1.9
Answer» 6.2x=(8.1)^2-(1.9)^2=(8.1+1.9)(8.1-1.9) 6 .2x=10×6.2 x=10×6.2/6.2 x=10 hope it is USEFUL for you

Discussion

42.	Mrs Rane gives 0.025 part of her salary to her son vinod every month . vinod , spends 0.8 part of his pocket money and saves the rest . if he saves rupees 277.50 in three months , how much does ,Mrs Rane earn a month ?
Answer» Mrs.Rane gives 0.025 part of her salary to het SON vinod every month . let's take Mrs.Rane salary to be x . So 0.025x is GIVEN to her son every month from which he only SAVES 20%of it . Money saved by him=0.025x(0.2)=0.005x. Money saved by him in three MONTHS =₹277.50.money saved by him in ONE month=₹92.50. So 0.005x=92.50=>x=92.50/0.005=₹18500.Mrs.Rane earns ₹18500 every month.

Discussion

43.	The inner diameter of a well is 4.20 metre and its depth is 10 metre. Find the inner surface area of the well. Find the cost of plastering it from inside at the rate Rs.52 per sq.m.
Answer» By APPLYING FORMULA $2\pi \times 2.1 \times 10 + 2\pi \times 2.1 {}^{2}$ SURFACE area=159.65 for COST =159.65×52=8302.09

Discussion

44.	a thin wire is in the shape of a circle of radius 77 cm it is bent into a square the side of the square is
Answer» $circumference = 2\pi \times r$ $2 \times 22 \div 7 \times 77$ 2×22×11=484cm SIDES of sides of square=perimeter ÷ 4 484÷4=121cm

Discussion

45.	ABCD is a square f is the midpoint of Ab and b is equal to 1 by 3 of BC if area of triangle FB is equal to 1 not 8 cm square find the length of AC
Answer» Wait i im writing on paper hey listen i here unable to SEND PDF FILE. can i send u somewhere ELSE

Discussion

46.	The average of first 50 natural numbers is …………. .
Answer» $\bold{Answer :}$ The first 50 natural numbers are 1, 2, 3, ..., 50 So, the 1ST term (a₁) = 1, the last term (a₅₀) = 50 Then, the sum of 1st N = 50 natural numbers be = n/2 * (a₁ + a₅₀) = 50/2 * (1 + 50) = 25 * 51 = 1275 Hence, the average of 1st 50 natural numbers be = 1275/50 = 25.5 # $\bold{MarkAsBrainliest}$

Discussion

47.	The circumference of the base of a cylinder is 3.2m,its height is 20m find its volume
Answer» PLEASE FIND the ANSWER below in the ATTACHMENT..

Discussion

48.	Solve x+1/x =25 1/25
Answer» GIVEN, $\frac{x+1}{x}=25\frac{1}{25}\\\\\frac{x+1}{x}=\frac{25\times25+1}{25}\\\\\frac{x+1}{x}=\frac{625+1}{25}\\\\\frac{x+1}{x}=\frac{626}{25}\\\\25(x+1)=626x\\\\25x+25=626x\\\\25=626x-25x\\\\25=601x\\\\x=\frac{25}{601}\\\\Note: a\frac{x}{y}=\frac{ay+x}{y}$

48.

Solve x+1/x =25 1/25

Answer»

GIVEN,

$\frac{x+1}{x}=25\frac{1}{25}\\\\\frac{x+1}{x}=\frac{25\times25+1}{25}\\\\\frac{x+1}{x}=\frac{625+1}{25}\\\\\frac{x+1}{x}=\frac{626}{25}\\\\25(x+1)=626x\\\\25x+25=626x\\\\25=626x-25x\\\\25=601x\\\\x=\frac{25}{601}\\\\Note: a\frac{x}{y}=\frac{ay+x}{y}$

Discussion

49.	A ladder of length 10 m is leaning against a wall making an angle of 60° with the horizontal plane. It touches the wall at point A. If the ladder is rearranged in a way that it leans against the wall making an angle 30° with the horizontal plane , it touches the wall at point B. Find the length of AB.please guys give me fast and full answer
Answer» Let the base of wall be point O and top be point D. Length of ladder(AD or BD) is 10 m. If we draw a right angled ∆AOD: angle DAO = 60°. Using value of cos 60° = 1/2. Cos 60° = AO/AD. Putting all the values we get 1/2 = AO / 10 . AO = 5m. Similarly in ∆BOD Cos 30° = OB/BD Cos 30° = √3/2. Putting all the values in the above equation: √3/2 = OB/10. OB = 5√3. Distance between A and B = OB - OA = 5√3 - 5 = 5(√3 - 1) = 3.66 m. Hope you understood my answer. If yes then PLZ MARK me as the brainliest.

Discussion

50.	A diamond falls and breaks into three pieces whose weights are in the ratio 3:4:5. the value of the diamond is proportionate to the square of its weight. if the value of the original diamond is rs.72, 000, what is the loss in value due to the breakage?
Answer» LOSS of 47000 RS MAY be ?

Discussion

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