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1.

After 17 years, Chelsea will be thrice as old as she is 5 years ago. Find her present age.1). 13 years2). 14 years3). 15 years4). 16 years

Answer»

Let her present age be ‘X’ YEARS

Then, 5 years AGO her age was (X - 5) and after 17 years her age WOULD be (X + 17)

⇒ (X + 17) = 3(X - 5)

⇒ X + 17 = 3X - 15

⇒ 2X = 17 + 15 = 32

⇒ X = 16

∴ Chelsea’s present age is 16 years
Discussion
2.

II) y2 = 40961). if x > y2). if x ≥ y3). if x < y4). x ≤ y

Answer»

From the given data,

⇒ $(x = {\rm{\;}}\SURD 1225)$ = ±35

Also given that y2 = 4096

⇒ y = √4096 = ± 64

When x = 35 and y = 64, then x < y

x = 35 and y = - 64, then x > y

x = - 35 and y = 64, then x < y

x = - 35 and y = - 64, then x > y

∴ RELATIONSHIP cannot be determined
Discussion
3.

II. b2 + 17b - 168 = 01). a < b2). a > b3). a ≤ b4). a ≥ b

Answer»

I. a2 - 42a + 597 = 256

⇒ a2 - 42a + 597 - 256 = 0

⇒ a2 - 42a + 341 = 0

⇒ a2 - 31a - 11a + 341 = 0

⇒ a(a - 31) - 11(a - 31) = 0

⇒ (a - 11)(a - 31) = 0

Then, a = +11 or a = +31

II. b2 + 17b - 168 = 0

⇒ b2 + 24b - 7B - 168 = 0

⇒ b(b + 24) - 7(b + 24) = 0

⇒ (b - 7)(b + 24) = 0

Then, b = +7or b = -24

So, when a = +11, a > b for b = +7 and a > b for b = -24

And when a = +31, a > b for b = +7 and a > b for b = -24

∴ So, we can OBSERVE that a > b.
Discussion
4.

Shubhada is having total Rs. 300 in coin of denomination Rs. 1, Rs. 2, Rs. 5. No. of 2 Rs. Coin is three times the no. of 5 Rs. Coins. Total no. of coins is 160. How many coins of Rs. 5 are with Shubhada?1). 352). 253). 204). 40

Answer»

Let the no. of Rs. 5 COINS be ‘x’

Let the No. of 1 Rs. COIN be ‘y’

No. of 2 Rs. Coin be ‘3x’

According to the question,

x + 3x + y = 160---- (1)

And, 5x + y + 6x = 300---- (2)

By simultaneously solving equations (1) & (2), we get,

x = 20, y = 80

∴ Number of Rs. 1 = 80 coins, Rs. 5 coins = 20 coins, Rs. 2 coins = 60
Discussion
5.

1). x > y2). x < y3). x ≥ y4). x ≤ y

Answer»

I. x2 – 0.21x + 0.002 = 0

⇒ x2 – 0.01x – 0.2x + 0.002 = 0

⇒ x(x – 0.01) – 0.2(x – 0.01) = 0

⇒ (x – 0.01) (x – 0.2) = 0

Thus, x = 0.01 or 0.2

II. y2 + 0.19y – 0.002 = 0

⇒ y2 + 0.2y – 0.01y – 0.002 = 0

⇒ y(y + 0.2) – 0.01(y + 0.2) = 0

⇒ (y + 0.2) (y – 0.01) = 0

Thus, y = – 0.2 or 0.01

So, when x = 0.01, for y = – 0.2, then x > y and when x = 0.01, for y = 0.01, then x = y

And when x = 0.2, for y = – 0.2, then x > y and when x = 0.2, for y = 0.01, then x > y

∴ We can observe that x ≥ y
Discussion
6.

1). a < b2). a > b3). a ≤ b4). a ≥ b

Answer»

$(\begin{array}{l}{\RM{I}}.{\rm{\;}}\frac{{286{a^2}}}{{15}} - 30a =- {\rm{\;}}\frac{{14{a^2}}}{{15}} - 18 + 9A\\ \Rightarrow \frac{{286{a^2}}}{{15}} + \frac{{14{a^2}}}{{15}} - 30a - 9a + 18 = 0\end{array})$ 

⇒ 20a2 - 39a + 18 = 0

⇒ 20a2 - 24A - 15A + 18 = 0

⇒ 4A(5a - 6) - 3(5a - 6) = 0

⇒ (4a - 3)(5a - 6)=0

Then, a = +3/4 = + 0.75 or a = +6/5 = +1.2

$(\begin{array}{l}{\rm{II}}.{b^2} - {\rm{\;}}\frac{{158b}}{{63}} =- {\rm{\;}}\frac{{11}}{7}\\ \Rightarrow {b^2} - {\rm{\;}}\frac{{158b}}{{63}} =- {\rm{\;}}\frac{{99}}{{63}}\end{array})$ 

⇒ 63b2 - 158b + 99 = 0

⇒ 63b2 - 81b - 77b + 99 = 0

⇒ 9b(7b - 9) - 11(7b - 9) = 0

⇒ (9b - 11)(7b - 9) = 0

Then, b = + 11/9 = +1.222 or b = +9/7 = +1.286

So, when a = +0.75, a < b for b = +1.222 and a < b for b = +1.286

And when a = +1.2, a < b for b = +1.222 and a < b for b = +1.286

∴ So, we can observe that a < b.
Discussion