2
Q:

# $\inline 2660 \: of\: \frac{1}{7}\:of\: \frac{1}{2}\: of\frac{1}{9.5}\: of\frac{1}{2}=?$

 A) 1 B) 10 C) 0.1 D) 100

Explanation:

$\inline 2660 \: of\: \frac{1}{7}\:of\: \frac{1}{2}\: of\frac{1}{9.5}\: of\frac{1}{2}= \frac{2660}{266}=10$

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.

6 114
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.

6 110
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

5 146
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

Explanation:

Total strength of the class is given by

15 + 44 + 4 + 3 - 1 = 65

6 222
Q:

An owner of a Dry fruits shop sold small packets of mixed nuts for Rs. 150 each and large packets for Rs. 250 each. One day he sold 5000 packets, for a total of Rs. 10.50 lakh. How many small packets were sold ?

 A) 2000 B) 3000 C) 2500 D) 3500

Explanation:

Let 's' be the number of small packets and 'b' the number of large packets sold on that day.

Therefore, s + b = 5000 ... eqn (1)

Each small packet was sold for Rs.150.
Therefore, 's' small packets would have fetched Rs.150s.

Each large packets was sold for Rs.250.
Therefore, 'b' large packets would have fetched Rs.250b.

Total value of sale = 150s + 250b = Rs. 10.5 Lakhs (Given)

Or 150s + 250b = 10,50,000 ... eqn (2)

Multiplying equation (1) by 150, we get 150s + 150b = 7,50,000 ... eqn (3)

Subtracting eqn (3) from eqn (2), we get 100b = 3,00,000
Or b = 3000

We know that s + b = 5000
So, s = 5000 - b = 5000 - 3000 = 2000.

2000 small packets were sold.