4
Q:

# The price of commodity X increases by 40 paise every year, while the price of commodity Y increases  by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity?

 A) 2010 B) 2011 C) 2012 D) 2013

Explanation:

Suppose commodity X wil cost 40 paise more than Y after Z  years. Then,

(4.20 + 0.40Z) - (6.30 + 0.15Z) = 0.40

$\inline&space;\fn_jvn&space;\Leftrightarrow$  0.25Z = 0.40 + 2.10  $\inline&space;\fn_jvn&space;\Leftrightarrow$  Z= $\inline&space;\fn_jvn&space;\frac{2.50}{0.25}&space;=\frac{250}{25}=10$

$\inline&space;\fn_jvn&space;\therefore$ X will cost 40 paise more than Y 10  years after 2001 . i.e in 2011

Q:

If two fractions, each of which has a value between 0 and 1, are multiplied together, the product will be :

 A) always greater than either of the original fractions B) always less than either of the original fractions C) sometimes greater and sometimes less than either of the original fractions D) remains the same

Answer & Explanation Answer: B) always less than either of the original fractions

Explanation:

We can easily Answer this by taking some values.

The question states that the two fractions, each have values between 0 and 1.

Let us say one of the fraction is 1/2 and the other fraction is 1/3 .

The product of the two fractions is 1/2 x 1/3 = 1/6 .
is lesser than both 1/2  and 1/3 .

So, the correct answer is that the product is always less than either of the original fractions

4 41
Q:

When 0.46 is written in simplest form, the sum of the numerator and the denominator is  ?

 A) 69 B) 73 C) 96 D) 41

Explanation:

0.46 = 46/100 = 23/50.

Sum of the numerator and denominator is 23 + 50 = 73

1 134
Q:

Evaluate : $\inline \fn_jvn \small \frac{3.39^{2}-2.61^{2}}{3.39-2.61}$ ?

 A) 3 B) 4 C) 6 D) 1

Explanation:

Given $\inline \fn_jvn \small \frac{3.39^{2}-2.61^{2}}{3.39-2.61}$ = $\inline \fn_jvn \small \frac{a^{2}-b^{2}}{a-b}$ = $\inline \fn_jvn \small \frac{(a+b)(a-b)}{a-b} = (a+b)$ = 3.39 + 2.61 = 6.

3 227
Q:

Karthik read $\inline \fn_jvn \small \frac{6}{13}$ th of a book in 1st week and $\inline \fn_jvn \small \frac{5}{9}$ th of the remaining book in 2nd week. If there were 100 pages unread after 2nd week, how many pages were there in the book ?

 A) 404 B) 415 C) 418 D) 420

Explanation:

Let the total book be 1
Then, in the 1st week $\inline \fn_jvn \small \frac{6}{13}$th of book is read
i.e remaining pages = $\inline \fn_jvn \small 1-\frac{6}{13} = \frac{7}{13}$
in 2nd week he read $\inline \fn_jvn \small \frac{5}{9}$th of it
i.e $\inline \fn_jvn \small \frac{5}{9}$ $\inline \fn_jvn \small \times \frac{7}{13} = \frac{35}{117}$
remaining pages after 2nd week = $\inline \fn_jvn \small \frac{7}{13}-\frac{35}{117} = \frac{28}{117}$
But given remaining pages after 2nd week = 100
$\fn_jvn&space;\small&space;\Rightarrow$ $\inline \fn_jvn \small \frac{28x}{117} = 100$
$\fn_jvn&space;\small&space;\Rightarrow$   x = 418

$\fn_jvn&space;\small&space;\therefore$ The total pages in the book are 418

6 290
Q:

$(0.2 * 0.2+0.01)(0.1 * 0.1+0.02)^{-1}$ is equal to

 A) 5/3 B) 9/5 C) 41/4 D) 41/12

Given expression = $\inline&space;\fn_jvn&space;\frac{(0.2\times&space;0.2+&space;0.01)}{(0.1\times&space;0.1&space;+0.02&space;)}=\frac{0.04+0.01}{0.01+0.02}=\frac{0.05}{0.03}=\frac{5}{3}$