4
Q:

# The Shopkeeper increased the price of a product by 25% so that customer finds it difficult to purchase the required amount. But somehow the customer managed to purchase only 70% of the required amount. What is the net difference  in the expenditure on that product?

 A) 10% more B) 5% more C) 12.5% more D) 17.5% more

Explanation:

Quantity  X  Rate =  Price

1  x  1  =  1

0.7 x 1.25 = 0.875

$\therefore$  Decrease in price = $\inline&space;\frac{0.125}{1}\times&space;100&space;=&space;12.5$ %

Q:

If three eighth of a number is 141. What will be the approximately value of 32.08% of this number ?

 A) 101 B) 112 C) 104 D) 120

Explanation:

K x 3/8 = 141 => K = 376

376 x 32.08/100 = 120

0 13
Q:

If L exceeds K by 30%, then K is less than L by ?

 A) 14 2/5 % B) 23 1/3 % C) 20 % D) 16 %

Explanation:

K = 100 L = 130
130------ 30
100------- ? => 23 1/13 %

2 87
Q:

A user type a content on 20 sheets, each sheet consist of 55 lines, and each line can have 66 characters. The content is again retyped on another set of sheets, these new sheets have 65 lines and each line can take 70 characters. Calculate the reduction in sheets ?

 A) 2 or 60% B) 4 or 20% C) 6 or 10% D) 8 or 5%

Explanation:

20 x 55 x 66 = L x 65 x 70
Here L = 72,600/4550 = 15.95 =~ 16
Reduction is 20-16 = 4

which is 20% reduction.

2 126
Q:

Kalyan purchased a Maruti van for Rs. 1,96,000 and the rate of depreciation is 14(2/7) % per annum. Find the value of the van after two years ?

 A) Rs. 1,24,000 B) Rs. 1,44,000 C) Rs. 1,32,000 D) Rs. 1,62,000

Explanation:

Value of Maruti Van, V1 = Rs. 1,96,000
Rate of depreciation, r = 14(2/7)% = 100/7%; Time, t = 2 years
Let V2 is the value after depreciation.
V2 = $\inline \fn_jvn \small V1[1-(r/100)]^{t}$
V2 = $\inline&space;\fn_jvn&space;\small&space;196000[1-((100/7)/100)]^{2}$
V2 = 196000(6/7)(6/7)
V2 = (196000x36)/49 = Rs. 1,44,000.

2 109
Q:

A duster is of dimension 5x5x7 all in cms,  is sinked into a cylindrical vessel of radius 5 cm and height 10 cm. After insertion the level of the water is inccresed by 2.14 cm. Then by what percentage the part of the duster is above the layer ?

 A) 96 % B) 4 % C) 97 % D) 3 %

Explanation:

Imension of the duster is assumed as 5x5x7 cm3

Volume of the duster below the water layer
= volume of water increased
= πxrxrxh = 3.14×52×2.14 ≈ 168 cm3

Total volume of the duster = 5×5×7 = 175 cm3

Percentage of the duster part below the layer = 168×100/175 = 96%

Percentage of the duster part above the layer = 4%