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Nachiket's investment was 38,000 rupees and share of his profit was 9,120 rupees.

Step-by-step explanation:

OK.So lets begin. . .

So Shweta, Piyush and Nachiket invested 80,000 rupees and started a business.

∴ Shweta & Piyush's total share is = (30,000+12,000) = 42,000 rupees.

∴ Nachiket's share should be = (80,000-42,000) = 38,000 rupees.

Now,

∴ Ratio of Shweta,Piyush and Nachiket's share = 30,000 : 12,000 : 38,000 = 15 : 6: 19

15+6+19 = 40.

∴ Profit of 24% on 80,000 rupees = 19,200 rupees.

∴ Nachiket's profit on 19,200 rupees was = 19200×19÷40 = 9,120 rupees. (19 is ratio of Nachiket's share and 40 is the sum of the share ratios of the three person).

thanks

sin=P/h cos=B/h √7225-1089=√6136 cos=√6136÷85

Mean age of 30 girls is 14 years.Hence, sum of ages of 30 girls= 30×14= 420.5 girls with mean age of 15 years have left.So, remaining sum = 420 - (15×5)= 420 -. 75= 345.Remaining no of girls = 30-5=25So, new mean age = Remaining sum/ Remaining girls= 345/25= 13.8 yearsHence, correct option is (D)

yes d is the correct answer

option D is used answer

(1/2)^-4 + (1/3)^-3-(1/4)^-2= 2^4+3^3-4^2= =16+21-16=21

(b) ka write answer 23 ho ga

a. (1/2)-4+(1/3)-3-(1/4)-2 (2)4+(3)3-(4)2 32+27-8 59-8 51

Ans. :- Option (2) is correct.

x^2+7x+10 = 0

x^2+5x+2x+10 = 0

x(x+5)+2(x+5) = 0

(x+5)(x+2) = 0

x = -2, -5

Number of girls/Number of boys = 7/15Number of girls/1200 = 7/15Number of girls = (7/15) × 1200 = 560

number of Girls's /number of boys =7/5no. of girls /1200=7/15no. of girls =(7/15)*1200=560

15x =1200X= 1200÷15=80girls=7×80=560

Thanks

thank you so much

what is your question

2.58 m of rope will be left

Let the first no. be x.

20 = x + (x-8) = x + x - 8 = 2x -8

20 + 8 = 2x 28 = 2x

28/2 = x14 = x

Hence, x = 14 & x - 8 = 6.

Ans : 14 &6

1260/4+5x+x1260/4+6x.......

the no is xother no,4+3x.........

let, 1st part be X, 2nd part be y, Given, x+y=1260....eqn. 1A\Q,x=3y+4putting X in eqn.1,x+y=1260or 3y+4+y=1260or 4y+4=1260or 4y=1256or y=314putting value of y in eqn. 1,x=1260-yor x=1260-314=946

and. 1st part is 946 2nd part is 314

let the second part be x them according to question❓first part = 4+3xnow, x + 4 + 3x = 1260 ( given ) 4x = 1256x = 314hence 1st part = 946and 2nd part = 314Accept as best 👍💯 👊💥

1st part is 945 and 2nd part is 315

X=946, Y=314 is right answer

Let the first part be=xSo,the second part be=x +18According to question:-38=x(x+18)38=x+x+1838=2x+1838-18=2x20=2x20/2=x10=xHence,first part=x =10Second part=x+18 =10+18 =28

10, 28 is the correct answer of the given question

1) perimeter of rectangle = 2(length + breath)= 2(3+2)=2×5=10cm2)perimeter = 2(5+2)= 2×7=14cn3)perimeter= 2(6+4)= 2×10=20cm

it is correct answer

perimeter= 2 ( l x b); 1) 2(3+2)=2(6)=12; 2) 2(5+3)=2(8)=16; 3)2(8+4)=2(12)=24

perimeter =2(lxb); 1) 2(3+2)=2(5)=10; 2) 2(5+3)=2(8)=16, 3) 2(8+4)=2(12)=24

it is correct answer

it is correct answer

it is correct answer

cot-¹(tanπ/7) = π/2 -tan-¹(tanπ/7) = π/2 -π/7 = (7-2)π/14 = 5π/14..

option C

thanks sister

volume of cuboid =lbh=2×3×5=30(b)volume =lbh=4×4×10=160

thanks alot

perimeter of the triangle= 3+5+7=15 cm

perimeter of a triangle=x+y+z=3+5+7=15

perimeter of this triangle = 3+5+7=15

3+5+7=15cm^2 is the perimeter of the scalene triangle of which sides are given in the question

15 is the correct answer h

perimeter of a triangle = s+s+ssides given = 3 cm, 5 cm, 7cmso 3+5+7= 15 cmtherefore perimeter of a triangle with sides 3 cm, 5cm, 7 cm is 15 cm

Perimeter of triangle= sum of all sides=3cm + 5cm + 7cm= 15cm

perimiter of triangle=3+5+7=15

answer = 153+5+7=15

the answer is 15cm because the formula is sum of all sides

perimeter=3+5+7=15cm

perimeter=3+5+7=15cm h

perimeter= 3+5+7=15cm answer.

15 cm is the correct answer

3+5+7=15 cm

so so easy

perimeter of traingle =Sum of all side 3cm +5cm +7cm=15cm

your image is Blur but I know the answer your answer is parameter equals to 5 cm of all sides equals to 37 + 5 + 7 cm equals to 8 cm and 7 CM 11 is your answer

10)Solution:-Let the average speed of the passenger train be 'x' km/hr and the average speed of the express train be (x + 11) km/hrDistance between Mysore and Bangalore = 132 kmIt is given that the time taken by the express train to cover the distance of 132 km is 1 hour less than the passenger train to cover the same distance.So, time taken by passenger train = 132/x hrThe time taken by the express train = {(132)/x+11} hrNow, according to the question{132/(x+11)} = 132/x + 1After taking L.C.M. of 132/x +1 and then solving it we get (132+x)/x.Now,{132/(x+11)} = (132+x)/xBy cross multiplying, we get132x = x²+132x+11x+1452x² +11x - 1452 = 0x² + 44x - 33x - 1452 = 0x(x+44) - 33(x+44) = 0(x+33) (x+44) = 0x - 44 or x = 33As the speed cannot be in negative therefore, x = 33 or the speed of the passenger train = 33 km/hr and the speed is 33 + 11 = 44 km/hrAnswer.

Let remaining two observations are x and y.We know, Mean = sum of observations/total number of observations ⇒6 = (5 + 7 + 9 + x + y)/5⇒6*5 = 21+ (x + y) ⇒ 30 - 21 = (x + y) ⇒ (x + y) = 9------(1)

Variance of given observations is 4we know, variance is given Variance = sum of (observation - mean) ^2 /total number of observations 4= {(5 - 6)² + (7 - 6)² + (9 - 6)² + (x - 6)² + (y - 6)²}/54*5 = (-1)² + (1)² + (3)² + x² + y² - 12(x + y) + 6² + 6² 20 = 1 + 1 + 9 + x² + y² - 12*9 + 36 + 36 20 = 83 + x² + y² - 108x² + y² = 128 - 83x² + y² = 45 ------(2)

Solve equations (1) and (2), x² + (9 - x)² = 45x² + 81 + x² - 18x = 452x² -18x + 36 = 0x² - 9x + 18 = 0 (x - 6)(x - 3) = 0x = 6, 3 put it in equation (1) Then, y = 9 - 6 = 3 or y = 9 - 3 = 6

Hence, remaining two observations are 3, 6