2
Q:

# Simplyfiy : $\inline \fn_cm \left [ 3\frac{1}{4}\div \left \{ 1\frac{1}{4}-\frac{1}{2}\left ( 2\frac{1}{2}-\overline{\frac{1}{4}+\frac{1}{6}} \right ) \right \} \right ]$

 A) 78 B) 88 C) 98 D) 108

Explanation:

Given exp =$\inline \fn_cm \left [ \frac{13}{4}\div \left \{ \frac{5}{4} -\frac{1}{2}\left ( \frac{5}{2}-\frac{3-2}{12} \right )\right \} \right ]$$\inline \fn_cm \left [ \frac{13}{4}\div \left \{ \frac{5}{4} -\frac{1}{2}\left ( \frac{5}{2}-\frac{1}{12} \right )\right \} \right ]$

$\inline \fn_cm \left [ \frac{13}{4}\div \left \{ \frac{5}{4} -\frac{1}{2}\left ( \frac{30-1}{2} \right )\right \} \right ]$ = $\inline \fn_cm \left [ \frac{13}{4}\div \left \{ \frac{5}{4} -\frac{29}{24}\right \} \right ]$ = $\inline \fn_cm \left [ \frac{13}{4}\div \left \{ \frac{30-29}{24} \right \} \right ]$=78

Q:

Given that 100.48 = x,  100.70 = y and xz = y2,  then the value of z is close to ___ ?

 A) 3.7 B) 2.7 C) 3.6 D) 2.9

Explanation:

xz = y2        10(0.48z) = 10(2 x 0.70) = 101.40

0.48z = 1.40

z = 140/18 = 35/12 = 2.9 (approx.)

4 72
Q:

What value will come in place of question mark (?) in the questions given below ?

[(180)2 ÷ 60×14] ÷ 9 = ? × 24

 A) 22 B) 33 C) 35 D) 28

Explanation:

Given

[(180)2 ÷ 60×14] ÷ 9 = ? × 24

[(180 x 180) ÷ 60×14] ÷ 9 = ? × 24

?= 180×3×14 / 9×24

? = 35.

6 309
Q:

The price of 2 oranges, 3 bananas and 4 apples is Rs. 15. The price of 3 oranges, 2 bananas and 1 apple is Rs. 10. What will be price of 4 oranges, 4 bananas and 4 apples  ?

 A) Rs. 10 B) Rs. 15 C) Rs. 20 D) Rs. 25

Explanation:

Let the Oranges be 'O', Bananas be 'B' and Apples be 'A'

From the given data,

2O + 3B + 4A = 15 ...... (I)

3O + 2B + A = 10 ..... (II)

5O + 5B + 5A = 25

O + B + A = 5

4O + 4B + 4A = 20.

6 479
Q:

Simplify the following :

$\inline \fn_jvn \frac{(37)^{3}+(35)^{3}+(28)^{3}-3x37x35x28}{(37)^{2}+(35)^{2}+(28)^{2}-37x35-35x28-37x28}$ ?

 A) 100 B) 1 C) 4 D) 0

Explanation:

Given $\inline \fn_jvn \frac{(37)^{3}+(35)^{3}+(28)^{3}-3x37x35x28}{(37)^{2}+(35)^{2}+(28)^{2}-37x35-35x28-37x28}$

It is in the form of

$\inline \fn_jvn \frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}-ab-bc-ca} = \frac{(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)}{a^{2}+b^{2}+c^{2}-ab-bc-ca}$

= a + b + c

Here a = 37, b = 35 & c = 28

=> a + b + c = 37 + 35 + 28 = 100

Therefore, $\inline \fn_jvn \frac{(37)^{3}+(35)^{3}+(28)^{3}-3x37x35x28}{(37)^{2}+(35)^{2}+(28)^{2}-37x35-35x28-37x28}$  = 100

5 335
Q:

In a slip test, K got the 15th rank and he was 44th from the bottom of the list of passed students. 4 students did not take up the slip test and 3 students were failed. What is the total strenght of the class ?

 A) 63 B) 62 C) 64 D) 65