IPMAT Rohtak 2020 Question Paper

Let us assume the present sum of money = M

In two years, the sum of money = 800

In another year (three years from beginning), the sum of money = 920

Therefore, in one year, the Interest of money added = Third year - 2nd year valuation = 920-800 = 120

Thus, the present sum of money = 800-2*120 = 560 

Let the amount invested in scheme A be 'P'. Hence, amount invested in scheme B would be '13900-P' .

In scheme A , as the interest rate is 14% per annum , for 2 years the interest = $$\dfrac{\ P\times\ 2\times\ 14}{100}$$ 
In scheme B, as the interest rate is 11% per annum, for 2 years the interest =  $$\left((13,900-P\right)\times\ 2\times\frac{\ 11}{100}$$

total interest = $$\dfrac{\ 22\left(13,900\right)-4P}{100}$$ = 3508 
Therefore, upon solving the above, we arrive at P = 7500.
Hence, amount invested in B = 13900-7500 = 6400

By seeing the options , we can infer that it is greater than $$316^2$$ . Let the value be 'a' which is above 316. Hence, lets rewrite the number :

$$\sqrt{99999}$$ = (316 + a)
99999 = $$316^2$$+ $$a^2$$ + 2(316)a
99999 = 99856 + $$a^2$$ + (632)a
Hence, we get : 143 =  $$a^2$$ + (632)a
For very small value of 'a' , $$a^2$$ can be negligible value , Hence a < $$\dfrac{\ 143}{632}$$ that will be a < 0.2262 .

Therefore 'a' should be less than 0.2262 as $$a^2$$ is present. Therefore the answer should be 316.22.

Given number is cube root of 55555 . By looking at the options we can infer that cube root of 55555 will be greater than 38 by a meagre amount of lets say 'x' :

55555 = $$(38+X)^3$$ 
55555 = 54872 +$$X^3$$ + 3(38)(X)(38+X)
683 = $$X^3$$ + 114(X)(38+X) 
      =  4332(X) +  114$$(X^2)$$ + $$X^3$$

"X" being a meagre value $$X^3$$ & 114$$(X^2)$$  doesn't contribute much to the 683 value . Hence X < $$\dfrac{\ 683}{4332}$$ .
Therefore X < 0.157.
Hence, the answer should be 38.15 .

Let the total no of products be 'P' . Now, as b1, b2, b3 are producing 20%, 15% and 32% :
b1  is producing = (0.2)P
b2 is producing = (0.15)P
b3 is producing = (0.32)P

As b1,b2,b3 produces 3%, 7% and 2% defective products respectively, it implies they make 97%, 93%, 98% products non-defective .
Hence, b1's no. of non defective products = 0.97(0.2)P = 0.194 P
b2's no of non defective products = 0.93 (0.15) P = 0.1395 P
b3's no of non defective products = 0.98 (0.32) P = 0.3136 P

Total non defective products = P (0.194 + 0.1395 + 0.3136) = 0.6471 P
Therefore, percentage of non defective products =  64.71 %

The question cannot be solved because of insufficient data .

If two persons moving with speeds of X kmph and Y kmph on a circular path of P km in the same direction from the common starting point, then they meet for first time after $$\dfrac{\ P}{\left|X-Y\right|}\ hrs$$.

In this question, P = 115 km; |X-Y| = |5-4| = 1 kmph

Therefore, they meet for first time after 115 hrs.

Hence, it is told that at least one racquet has to be there in each identical boxes, the only possibilities to arrange 5 different racquets will be : (1,1,3) or (1,2,2 ) .

Case I : (1,1,3) = Selecting 3 racquets from 5 different racquets and rest of the two goes one in each box = $$5_{C_3}$$ = 10 ways
Case II : (1,2,2) = Selecting 4 racquets from 5 different racquets and arranging them in (2,2) way in 2 identical                                             boxes  = $$5_{C_4}\times\dfrac{\ 4_{C_2}}{2}$$ = 15 ways 

Therefore, the total arrangements = 15 + 10 = 25

To make a signpost either we can use 1 symbol or 2 symbols or 3 symbols or 4 symbols or 5 symbols or 6 symbols . When we use "X" particular no of symbols of different colours we can arrange them in $$X!$$ .

Hence total ways = $$6_{P_1}+6_{P_2}+6_{P_3}+6_{P_4}+6_{P_5}+6_{P_6}$$
                           = 1956.

Copies of Beetle CD's = 12
Copies of Pink Floyd CD's  = 7
Different CD's of MJ = 3
Different DC's of Madonna = 2 

In the case of 5 (3+2) different CD's of MJ & Madonna, each CD has two possibilities that is either be selected or not selected . Hence for different CD's it would be $$2^3\times2^2$$ ways.

In case of copies of Beetle CD's either 0 can be selected or 1 can be selected or 2 can be selected & so on.. till 12 i.e {0,1,2,3...12} . Hence for this no of ways will be 1+1+1+1....13 times = 13 ways.

In case of copies of Pink Floyd CD's either 0 can be selected or 1 can be selected or 2 can be selected & so on.. till 7 i.e {0,1,2,3...7} . Hence for this no of ways will be 1+1+1+1....8 times = 8 ways

Total no of ways of selection = $$2^5$$x13x8 = 3328. 
In the no of ways calculated above, there is one way in which none of them are selected , but the question is about atleast 1 or more to be selected , Hence subtract 1 from above number i.e 3327 is no of ways in which one or more can be selected.

Therefore, the answer is 3327.