Algebra Question and Answers Quiz-1

Option(B) is correct

Let the age be x and y years now.

Then, x−y=20 -------- (i)

and (x−5)=5(y−5) ------- (ii)

On solving both equation we get:

x=30 and y=10

Total Debt = 3600

Amount Paid at Death = \frac{2}{3} \times 3600 = 2400

The total debt was to be paid in 40 installments.

Using formula of AP,

Sum of n terms of an AP = \frac{n}{2} \times (First term + Last term)

3600 = \frac{40}{2} \times (a + T40)

180 = a + T40 ....... (1)

Now 2400 was paid in 30 installments

2400 = 30/2 (a + T30)

160 = a + T30 ...... (eqn 2)

Subtract eqn 2 from eqn 1

20 = T40 - T30

Nth term of  AP: Tn = a + (n-1)d

20 = a + 39d - (a + 29d)

20 = 10d

2 = d

substitute in eqn 1, we get a = 51

Solution:
Let the number of students be x. Then,

\frac{x}{2} - 23\sqrt{x} = 256

x - 56\sqrt{x} = 512

x - 56\sqrt{x} - 512 = 0

Let \sqrt{x} = y

y^{2} - 56y - 512 = 0

On solving the quadratic equation, we get

(y - 64)(y + 8) = 0

y = 64 , y = -8

Thus y = 64

Thus, \sqrt{x} = 64

x = 64^{2} = 4096

Solution:

Let the number be x.

As per what he did, \frac{x+12}{6} = 112

x+12 = 672

x = 672-12 = 660.

So, the correct answer = \frac{660}{6}+ 12 = 110+12 = 122.

Let the cost of potatoes per kg = Rs. x

And the cost of tomatoes per kg = Rs. y

6x + 2y = 40 ........ (i)

2x + 6y = 60 ........ (ii)

Solving equation 1 and 2, we get
x = 3 and y = 11

Solution:
Option(B) is correct

Machines A and B together will produce 3,00,000 m of cloth in 10 hour.

Machines A and B together will produce \frac{3,00,000}{10} = 30,000 m of cloth in 1 hour.

Machine B alone can produce 3,00,000 m cloth in 15 hour.

Machine B alone can produce \frac{30,000}{15} = 20,000 m cloth in 1 hour.

Therefore, Machine A can produce 30,000 - 20,000 = 10,000 m cloth in 1 hour.

So, in 10 hour Machine A can produce 10,000 \times 10 = 1,00,000 m of cloth.

Solution:

Let A's age be x and B's age be y.

x+y=43 -------- (i)

(x+11) = 76 \times (y+11)

⇒6x−7y=11 -------- (ii)

On solving both the equations, we get:

y = 19 years

Solution:
B + 8 = C....(i)

A − 8 = C – 3....(ii)

A + 6 = 2D......(iii)

B + D = 50......(iv)

Substitute (i) in (ii)

A - B = 13....(v)

From (iii), get expression of D = \frac{A +6}{2}, and substitute in (iv)

A + 2B = 94....(vi)

solving (v) & (vi), we get A=40

Let age of Rishabh in 2002 = x

So, 2002 − 90x = x

⇒x=22

So, Rishabh’s age in 2006 is,

=> 22 + 4 = 26 years

here it is given that

x + \frac{1}{x} = 2

and it is possible only when x = 1 and hence we will put x = 1 in

x^{10} + \frac{1}{x^{10}} = 1 + 1 = 2