IPM Indore 2019 Question Paper

The vertices of the triangle are (0,0), (31,0), and (0,31). Thus, this is a right-angled triangle, whose equation of hypotenuse is given by $$ x+y=31$$. Now, we need to find the points whose both coordinates are integers, and lie inside the triangle. 

This means that - $$x>0$$, $$y>0$$, and $$x+y<31$$.

$$x+y<31$$ => $$y<31-x$$ and $$y>0$$ => $$0<y<31-x$$

If x = 0, then y can take 30 values {1,2,3,...,30}
If x = 1, then y can take 29 values {1,2,3,...,29}
If x = 2, then y can take 28 values {1,2,3,...,28}
.
.
.
If x = 29, then y can take only one value {1}

Thus, the total number of points inside the triangle = $$1+2+3+.....+30$$ = $$\dfrac{30\times 31}{2}=435$$ points. 

Let the speed of two small insects be v1 and v2, where v1 > v2.

When they are travelling towards each other, then using the relative motion concept, their speeds will add up, and it takes u minutes to cover x metres distance

=> $$v1+v2=\dfrac{x}{u}$$

When they are travelling in the same direction, then using the relative motion concept, the difference of the speeds will be taken, and it takes v minutes to cover x metres distance

=> $$v1-v2=\dfrac{x}{v}$$

Dividing both equations -

$$\dfrac{v1+v2}{v1-v2}=\dfrac{v}{u}$$

Applying Componendo-Dividendo

$$\dfrac{\left(v1+v2\right)+\left(v1-v2\right)}{\left(v1+v2\right)-\left(v1-v2\right)}=\dfrac{v+u}{v-u}$$

$$\dfrac{2v1}{2v2}=\dfrac{v1}{v2}=\dfrac{v+u}{v-u}$$

The ratio of the speed of the slower insect to the faster insect is = $$\dfrac{v2}{v1}=\dfrac{v-u}{v+u}$$

An alloy P has copper and zinc in the proportion of 5:2 (by weight), while another alloy Q has the same metals in the proportion of 3:4 (by weight).

Alloy P and Alloy Q are mixed in the ratio a:b to form alloy R. Let us assume the amount of alloy P is 7a, and the amount of alloy Q is 7b. 

The amount of Copper and Zinc in alloy P will be 5a and 2a, respectively. The amount of Copper and Zinc in alloy Q will be 3b and 4b, respectively. Thus, in alloy R, the amount of Copper and Zinc will be 5a+3b and 2a+4b, respectively. And it is given that alloy R has equal contents of copper and zinc.

=> 5a + 3b = 2a + 4b => b = 3a => a:b = 1:3.

Now, alloy S is formed by mixing the two alloys P and Q in the proportion of b:a. That is, they are mixed in a 3:1 ratio. Let us assume the amount of alloy P as 21x, and the amount of alloy Q as 7x.

The amount of Copper and Zinc in alloy P will be 15x and 6x, respectively. The amount of Copper and Zinc in alloy Q will be 3x and 4x, respectively. Thus, in alloy S, the amount of Copper and Zinc will be 15x + 3x = 18x and 6x + 4x = 10x, respectively.

Thus, the proportion of copper and zinc in the alloy S will be 18x:10x = 9:5.

For any number which has less than 7 digits, each place in the number can be filled in two ways [either 1 or 3].

Total number of 1-digit numbers = 2

Total number of 2-digit numbers = 2 x 2 = 4

Total number of 3-digit numbers = 2 x 2 x 2 = 8

Total number of 4-digit numbers = 2 x 2 x 2 x 2 = 16

Total number of 5-digit numbers = 2 x 2 x 2 x 2 x 2 = 32

Total number of 6-digit numbers = 2 x 2 x 2 x 2 x 2 x 2 = 64

For 7-digit numbers, we need the number to be less than 3000000, thus the first digit can only be, and the rest digits can be filled in 2 ways. Thus, 

Total number of 7-digit numbers = 1 x 2 x 2 x 2 x 2 x 2 x 2 = 64

Total numbers = 2 + 4 + 8 + 16 + 32 + 64 + 64 = 190

The possible products for a pair of integers, all of which are chosen from either $$+1$$ or $$-1$$, are:

• $$+1$$ and $$+1$$ having the product $$+1$$
• $$+1$$ and $$-1$$ having the product $$-1$$
• $$-1$$ and $$-1$$ having the product $$+1$$

Therefore, the only way that many such pairs have the sum of their products as $$0$$ is that there are as many products of $$+1$$s as there are $$-1$$s. From $$a_1a_2$$ to $$a_{n}a_1$$ there are a total of $$n$$ pairs. Since these $$n$$ number of pairs have $$\frac{n}{2}$$ pairs having the product $$+1$$ and $$\frac{n}{2}$$ pairs having the product $$-1$$, we can definitively say that $$n$$ has to be divisible by $$2$$ for $$\frac{n}{2}$$ to be an integer. 

Furthermore, the pairs are constructed cyclicly, we start with $$a_1$$ and end in $$a_1$$. Each term, therefore, affects the value of another. There have to be $$\frac{n}{2}$$ pairs that have the product $$-1$$, and these have the two terms as $$+1$$ and $$-1$$ in some order. There will also be $$\frac{n}{2}$$ pairs having the product $$+1$$, since these products are also given by adjacent terms, the terms having the values $$+1$$ and $$+1$$ will occur in pairs and those having the value $$-1$$ and $$-1$$ will occur in pairs. The number of terms, therefore, will further be divisible by 2. 

In essence:

• There is the same number of pairs having the product $$+1$$ as the number of those that have the product $$-1$$.
• For each product $$+1$$, the adjacent terms occur in pairs of $$(+1,+1)$$ or $$(-1,-1)$$; these are then arranged just so, so that there are adjacent products of $$-1$$ possible.

Therefore, $$n$$ should be twice-divisible by $$2$$, or that $$n$$ should be divisible by $$4$$. $$n$$ can therefore be any multiple of $$4$$.

Average = Sum of Annual exports across all year / Total number of years 

=> Average = $$\dfrac{280+280+230+210+200+220+210+200+200+200+220+200}{12}$$

=> Average = $$\dfrac{2650}{12}$$

=> => Average = $$220.83$$ Crores

Percentage decline during the 2006 to 2011 period = (Difference between exports in 2006 and 2011 / 2006 exports) * 100

Percentage decline during the 2006 to 2011 period = $$\dfrac{280-220}{280}\times100=21.42\%$$

Percentage decline during the 2012 to 2017 period = (Difference between exports in 2012 and 2017 / 2012 exports) * 100

Percentage decline during the 2012 to 2017 period = $$\dfrac{210-200}{210}\times100=4.76\%$$

Thus, the first percentage is more than the second percentage by $$21.42-4.76=16.66$$, which is approximately $$16.5$$. 

Maximum Difference = 120 Cr. [ in 2014 ]

2007-2009
2007: 280 - 240 = 40
2008: 230 - 260 = -30
2009: 210 - 280 = -70
Average = (40 - 30 - 70)/3 = -20

2015 - 2017
2015: 200 - 310 = -110
2016: 220 - 250 = -30
2017: 200 - 240 = -40
Average = (-30 - 40 - 110)/2 = -60

2014 - 2016
2014: 200 - 320 = -120
2015: 200 - 310 = -110
2016: 220 - 250 = -30
Average = (-120 - 110 - 30)/3 = -86.67

2010 - 2012
2010: 200 - 310 = -110
2011: 220 - 290 = -70
2012: 210 - 320 = -110
Average = (-110 - 70 - 110)/3 = -96.67

Largest average negative balance of trade = -96.67, for the year 2010 - 2012. 

2009 - 2010
%age change = [(310 - 280)/280] * 100 = 10.71%

2010 - 2011
%age change = [(290 - 310)/310] * 100 = -6.45%

2013 - 2014
%age change = [(320 - 300)/300] * 100 = 6.67%

2008 - 2009
%age change = [(210 - 230)/230] * 100 = -8.69%

Percentage increase is maximum in 2009 to 2010.