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Answer: AC³ > AB³ + BC³

Explanation:

Let’s take an example –

5,4,3 5³ = 125 4³ + 3³ = 64 + 27 = 91

So it’s clear that AC³ > AB³ + BC³

Answer-: 18 km/hr

Solution-:

D. S. T.

50 25. 2

40. 20. 2

90. 15. 6

Average speed = total distance/total time

= (50+40+90)/(2+2+6)

= 180/10 = 18 km/h

Answer-: 23 days

S + R —> 18 days

R + N —> 24 days

S + N —-> 36 days

Let total work = 72 units

Efficiency of ,

S+R –> 4 units/day

R+N –> 3 units/day

S+N –> 2 units/day

We can say , S –> 1½ units/day

R –> 2½ units/day

N –> ½ units/day

Work done by all in 2 days = 2×(1½+2½+½) = 9 units

Ramaining work done by Rohan and Neeta in = (72-9)/(2½+½) = 21 days

Total days = 21+2 = 23 days

Let the distance b/w two ports be X km.

Then, speed downstream = X/3

speed upstream = X/4

Now,

Speed of stream = ½(X/3 –X/4)

5 = ½×X/12

X = 120 km

Ans –: angle BDC = 44°

SOLUTION –:

Angle ADB = 180°-59° = 121°

By exterio angle property of triangle

4x + 7x = 121°

x = 11

Angle BDC = 4×11 = 44°

ANS –: 52,000/-

SOL –:

16.67% —-> 1/6

Madhu. Pooja

7x 6x —income

6y 5y —expenditure

16000. 14000 —- saving

So, we get

7x – 6y = 16000—-eq¹

6x – 5y = 14000 —–eq²

By solving eq¹ & eq² we get x = 4000

So , total income of Madhu & Pooja

= 13×4000 = 52,000 /-

Solution -:

1/(1+ sinA) + 1/(1-sinA) = (1-sinA+ 1+sinA)/cos²A = 2/cos²A = 2sec²A

answer -: (4). Rs. 20400/-

solution-:

27. Let total work = 600 A does per day = 600/40 = 15 B = 600/50 = 12 C = 600/60 = 10

Let work completed in X days So 15*5 + 12*(X-3) + 10*X= 600 22X = 561 X (total days) = 25.5 days

Work done by C = 10*25.5 = 255

Share of C = 48000*(255/600) =Rs. 20400

ans. (1) 79

sol-

The average of 25 consecutive odd integers = 55

So, we can say middle term ( 13th no.)

= 55

So, 25th no. = 55 + 2×12 = 79

## Aakashmishra

Answer: AC³ > AB³ + BC³

Explanation:

Let’s take an example –

5,4,3

5³ = 125

4³ + 3³ = 64 + 27 = 91

So it’s clear that AC³ > AB³ + BC³

## Aakashmishra

Answer-: 18 km/hrSolution-:D. S. T.

50 25. 2

40. 20. 2

90. 15. 6

Average speed = total distance/total time

= (50+40+90)/(2+2+6)

= 180/10 = 18 km/h

## Aakashmishra

Answer-: 23 daysSolution-:S + R —> 18 days

R + N —> 24 days

S + N —-> 36 days

Let total work = 72 units

Efficiency of ,

S+R –> 4 units/day

R+N –> 3 units/day

S+N –> 2 units/day

We can say , S –> 1½ units/day

R –> 2½ units/day

N –> ½ units/day

Work done by all in 2 days = 2×(1½+2½+½) = 9 units

Ramaining work done by Rohan and Neeta in = (72-9)/(2½+½) = 21 days

Total days = 21+2 = 23 days

## Aakashmishra

Solution-:Let the distance b/w two ports be X km.

Then, speed downstream = X/3

speed upstream = X/4

Now,

Speed of stream = ½(X/3 –X/4)

5 = ½×X/12

X = 120 km

## Aakashmishra

Ans –: angle BDC = 44°SOLUTION –:

Angle ADB = 180°-59° = 121°

By exterio angle property of triangle

4x + 7x = 121°

x = 11

Angle BDC = 4×11 = 44°

## Aakashmishra

ANS –: 52,000/-SOL –:16.67% —-> 1/6

Madhu. Pooja

7x 6x —income

6y 5y —expenditure

16000. 14000 —- saving

So, we get

7x – 6y = 16000—-eq¹

6x – 5y = 14000 —–eq²

By solving eq¹ & eq² we get x = 4000

So , total income of Madhu & Pooja

= 13×4000 = 52,000 /-

## Aakashmishra

Solution -:1/(1+ sinA) + 1/(1-sinA)

= (1-sinA+ 1+sinA)/cos²A

= 2/cos²A

= 2sec²A

## Aakashmishra

answer -: (4). Rs. 20400/-solution-:27.

Let total work = 600

A does per day = 600/40 = 15

B = 600/50 = 12

C = 600/60 = 10

Let work completed in X days

So

15*5 + 12*(X-3) + 10*X= 600

22X = 561

X (total days) = 25.5 days

Work done by C = 10*25.5 = 255

Share of C = 48000*(255/600) =Rs. 20400

## Aakashmishra

ans. (1) 79sol-The average of 25 consecutive odd integers = 55

So, we can say middle term ( 13th no.)

= 55

So, 25th no. = 55 + 2×12 = 79