MAH CET 2018 exam is scheduled to take place on March 10^{th} and 11^{th}, 2018. Only a few days are left for the preparation. So, the aspirants must not leave even a single stone unturned in order to get admission in MAH-CET top institute. Keeping this aspect in focus, we have prepared a practice set that will aid you in your preparation of the exam.

The Practice Set is on the ‘Quantitative Aptitude Section’ and comprises of 10 questions each from various topics.

**1. Ram packs sweets in such a way that if he packs 20 in each box, there would be 15 sweets left. If he packs 25 each, there would be 20 sweets left, if he packs 38 each, there would be 33 sweets left and if he packs 40 sweets in each box, there would be 35 sweets left. What is the minimum number of sweets the person have?**

a) 3795

b) 4125

c) 4250

d) 3985

e) None of these

**2. Sum of square of first number and cube of second number together is 988.Also square of the second number is 19 less than the square of 10.What will be the value of five times of first number.(assuming both the numbers are positive )**

a) 70

b) 65

c) 75

d) 80

e) 90

**3.**

**MH-CET Quantitative Aptitude Practice Questions Set-2**

**4. The average marks of 5 boys are decreased by 3 when one of them having 150 marks is replaced by another boy. This new boy is again replaced by another boy whose marks are 30 less than the boy replaced. What is the overall change in the average due to this dual change? **

a) 9

b) 15

c) 12

d) 6

e) None of these

**5. In an organization, in the company’s books of accounts, depreciation is charged at the rate of 10% of the previous value. However, every second year there is some maintenance work so that in that particular, depreciation is only 5% of its previous value. If at the end of fourth year, the value of the machine stands at Rs. 2, 92,410 then find the value of machine at the beginning of the first year.**

a) 3,00,000

b) 5,00,000

c) 4,00,000

d) 4,50,000

e) None of these

**6. If the dice are thrown simultaneously then the probability of sum appearing on the dice is less than 3?**

a) 5/11

b) 6/13

c) 1/36

d) 1/18

e) 1/12

**7. The area of the three adjacent faces of a rectangular box which meet in a point are known the product of these areas is equal to:**

a) The volume of the box

b) Twice the volume of the box

c) The square of the volume of the box

d) The cube root of the volume of the box

e) None of these

**8. ABCD is a square, 4 equal circles are just touching each other whose centres are the verticals A, B, C and D, of the square. What is the shaded area if the side of square is 2cm? **

**Directions (9 - 10) **Out of the two bar graphs provided below, one show the total number of candidates appeared in an examination over the years and the other shows the number of candidates selected over the years.

** Number of candidates appeared over the year**

**9. During which of the following year the difference between the numbers of candidates appeared and selected was maximum?**

a) 2005

b) 2009

c) 2007

d) 2010

e) 2006

**10. What was the difference between the average number of appeared candidates and average number of selected candidates during this period (in hundreds)? **

a) 72.5

b) 76.2

c) 77.5

d) 70.5

e) 69.5

Question |
Answer |

1. |
a |

2. |
b |

3. |
a |

4. |
a |

5. |
c |

6. |
e |

7. |
c |

8. |
b |

9. |
c |

10. |
c |

**Explanation: 1.** Number of sweets in each box =LCM (20, 25, 28, 38 and 40)–5

3,800−5= 3,795.**Explanation: 2.** Let the first number be X and second number be Y.

According to question,

X^{2} + Y^{2} = 998

And, Y^{2} = 100 – 19

Y = 9, X = 13 and 5 times = 65

Category: Equations**Explanation: 3. **

**Cubing both sides,z**

**Explanation: 4.** The marks of second boy are 135 and that of the third is 105. Hence, net result is a drop of 45 for 5 boys. Hence, the net change is of 9 marks.**Explanation: 5.** Let the price of machine at the beginning of the first year be Rs. X. According to question,

2, 92, 410 = X × 90/100×95/100×90/100×95/100

X = 4, 00,000.**Explanation: 6.** Total number of sum that can appear on two dice are (1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6) ,…………………………………………… (6, 6).

So , total number of possible events = 6× 6 = 36

Total sum appearing less than 4 is (1,1), (1,2), (2,1)

Number of favourable events = 3

Required probability = (Number of favourable events)/(Number of total possible events)

= 3/36 = 1/12

**Explanation: 7.** Let length = l, breadth = b and height = h. Then,

Product of areas of 3 adjacent faces = (lb × bh × lh) = (lbh)^{2} = (volume)^{2}

**Explanation: 8.** Side of the square = 2 cm means radius of the circle = 1 cm

Area of the shaded region = Area of square - Area of four quadrants

Area of the shaded region = (2 × 2) – (1/4 × 4 × πr^{2})

= 4 – π **Explanation: 9.**

The difference between the numbers of candidates appeared and selected

For 2005 = (200 - 120) l = 80.

For 2006 = (300 - 225) = 75

For 2007 = (500 - 375) = 125

For 2008 = (400 - 330) = 70.

For 2009 = (600 - 525) = 75.

For 2010 = (460 - 420) = 40.

Clearly, maximum difference was during 2007.

**Explanation: 10.** Average of appeared candidates = 1/6 (200 + 300 + 500 + 400 + 600 + 460) = 410

Average of selected candidates =1/6 (120 + 225 + 375 + 330 + 525 + 420) = 332.5

Required difference = (410 - 332.5) = 77.5

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