**Question 1**

A manufacturer reduces the size of his machine of Material1 from 30" to 28" and that of Material2 from 24" to 23". He normally spends Rs.1440 for Material1 and Material2. 3/4th of the amount spent would be for Material1. How much will he save under the size of new size (Assuming that the costs incurred by the machines are directly proportional to their sizes)?

a)Rs.85 b)Rs.87 c)Rs.86 d)Rs.88

**Answer :**b)Rs.87

Solution:

The manufacturer totally spends Rs.1440 for both material 1 and 2.

Out of this 3/4th is entirely for Material1.

Amount spent on Material1 = 1440 x 3/4 = Rs.1080

i.e, he spent Rs.1080 for Material1 of 30" and he reduces 2".

Size Amount 30" Rs.1080 2" ?Amount saved on Material1 = 1080 x 2/30 = Rs.72

Amount spent on Material2 = total amount - amount for Material1 = 1440 - 1080 = Rs.360.

i.e., he spent Rs.360 for Material2 of 24" and he reduces 1".

Size Amount 24" Rs.360 1" ?Amount saved on Material2 = 360 x 1/24 = Rs.15

Therefore total amount saved after size reduction = 72 + 15 = Rs.87

**Question 2**

The dimensions of a certain machine are X" x Y" x Z". If the average of its dimensions equals 680" and none of the dimensions is less than 640", what is the greatest possible length of one of the dimensions?

a)760" b)800" c)720" d)745"

**Answer :**a) 760"

Solution:

Given (X+Y+Z)/3 = 680".

Therefore the total length of the dimensions is X + Y + Z = 3 x 680 = 2040.

Let us assume Z is the dimension with greatest possible length.

From above equation, Z = 2040 - X - Y ...(1)

Based on eq (1), for Z to be of maximum value, X and Y should be as low as possible. Since it is given that none of the dimensions is lesser than 640", the lowermost value that X and Y can get is 640".

Substituting X = Y = 640" in eq (1), we get,

Z = 2040 - 640 -640 = 760".

**Question 3**

What will be the ratio of the shortest side of new dimensions to the greatest side of the old dimensions if the size of a certain machine is increased proportionally until the sum of its dimensions equals 800" and the dimensions of old size are 130" x 60" x 360" ?.

a)1:2 b)3:5 c)5:18 d)18:3

**Answer :**c) 5:18

Solution:

From the given data, we have

The dimensions of the machine before size increment = 130" x 70" x 360".

Sum of the dimensions = 130+70+360 = 560"

Note that the greatest side = 360" & smallest side = 70". ...(A)

It is given that the sum of the dimensions of new size = 800"

Since dimensions are increased proportionately, the smallest side of the old dimensions before increment will be the smallest side of new dimension after size increment.

Sum Side 560" 70" 800" ? Smallest side will be increased to 800 x 70 / 560 = 100" i.e., the smallest side of the new dimensions is 100". We have already found that the the greatest side of old dimensions is 360". Required Ratio = Smallest side of the New Dimensions : Greatest side of the Old Dimensions = 100 : 360 = 5 : 18 Hence the answer is 5 : 18.

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