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Q:

At a casino in mumbai, there are three tables A, B and C. The payoffs at A is 10:1, at B is 20:1 and at C is 30:1. If a man bets Rs 200 at each table and win at two of the tables, what is the maximum and minimum difference between his earnings can be?

A) Rs.2500	B) Rs.2000
C) Rs.4000	D) None of these

Answer & Explanation Answer: C) Rs.4000

Explanation:

Maximum earning will be only when he will won on the maximum yielding table.

A ----> 10:1

B ----> 20:1

C ----> 30:1

i.e, he won on B and C but lost on A

20 x 200 + 30 x 200 -1 x 200 = 9800

minimum earning will be when he won on table A and B and lose on that table 3.

Therefore, 10 x 200 + 20 x 200 - 1 x 200 = 6000-200 = 5800

Therefore, Difference= 9800 - 5800 = 4000

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Filed Under: Ratios and Proportions

Q:

The ratio of students in a coaching preparing for B.tech and MBA is 4 : 5. The ratio of fees collected from each of B.tech and MBA students is 25 : 16. If the total amount collected from all the students is 1.62 lakh, what is the total amount collected from omly MBA aspirants?

A) Rs. 62,000	B) Rs. 72,000
C) Rs. 82,000	D) None of these

Answer & Explanation Answer: B) Rs. 72,000

Explanation:

The ratio of fees collected from B.Tech : MBA = 4x * 25y : 5x * 16y

= 100xy : 80xy

= 5xy : 4 xy = 5k : 4k

The amount collected only from MBA students = $\frac{4}{9} \times 1.62 l a k h$ = Rs. 72,000

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Filed Under: Ratios and Proportions

Q:

The concentration of petrol in three different mixtures (petrol and kerosene) is 1/2 , 3/5 and 4/5 respectively. If 2 litres, 3 litres and 1 litre are taken from these three different vessels and mixed. what is the ratio of petrol and kerosene in the new mixture?

A) 4:5	B) 3:2
C) 3:5	D) 2:3

Answer & Explanation Answer: B) 3:2

Explanation:

Concentration of petrol in A B C

1/2 3/5 4/5

Quantity of petrol taken from A = 1 litre out of 2 litre

Quantity of petrol taken from B = 1.8litre out of 3 litre

Quantity of petrol taken from C = 0.8 litre out of 1 litre

Therefore, total petrol taken out from A, B and C = 1+1.8+0.8 =3.6 litres

So, the quantity of kerosen =(2+3+1) - 3.6 =2.4 litre

Thus, the ratio of petrol to kerosene = 3.6/2.4 = 3/2

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Filed Under: Ratios and Proportions

Q:

A child has three different kinds of chocolates costing Rs.2, Rs.5, Rs.10. He spends total Rs. 120 on the chocolates. what is the minimum possible number of chocolates he can buy, if there must be atleast one chocolate of each kind?

A) 22	B) 19
C) 17	D) 15

Answer & Explanation Answer: C) 17

Explanation:

Minimum number of chocolates are possible when he purchases maximum number of costliest chocolates.

Thus, 2 x 5 + 5 x 2 =Rs.20

Now Rs.100 must be spend on 10 chocolates as 100 = 10 x 10.

Thus minumum number of chocolates = 5 + 2 + 10 = 17

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Filed Under: Ratios and Proportions

Q:

The H.C.F and L.C.M of two numbers are 11 and 385 respectively. If one number lies between 75 and 125 , then that number is

A) 77	B) 88
C) 99	D) 110

Answer & Explanation Answer: A) 77

Explanation:

Product of numbers = 11 x 385 = 4235

Let the numbers be 11a and 11b . Then , 11a x 11b = 4235 => ab = 35

Now, co-primes with product 35 are (1,35) and (5,7)

So, the numbers are ( 11 x 1, 11 x 35) and (11 x 5, 11 x 7)

Since one number lies 75 and 125, the suitable pair is (55,77)

Hence , required number = 77

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Filed Under: HCF and LCM

Q:

If the sum of two numbers is 55 and the H.C.F and L.C.M of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to

A) 55/601	B) 601/55
C) 11/120	D) 120/11

Answer & Explanation Answer: C) 11/120

Explanation:

Let the numbers be a and b . Then, a+b =55 and ab = 5 x 120 = 600.

Therefore, Required sum = $\frac{1}{a} + \frac{1}{b} = \frac{a + b}{a b} = \frac{55}{600} = \frac{11}{120}$

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Filed Under: HCF and LCM

Q:

The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 100, then their difference is

A) 10	B) 46
C) 70	D) 90

Answer & Explanation Answer: A) 10

Explanation:

Let the numbers be x and (100-x).

Then, $x (100 - x) = 5 * 495$

=> $x^{2} - 100 x + 2475 = 0$

=> (x-55) (x-45) = 0

=> x = 55 or x = 45

The numbers are 45 and 55

Required difference = (55-45) = 10

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Filed Under: HCF and LCM

Q:

The H.C.F and L.C.M of two numbers are 84 and 21 respectively. If the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is

A) 12	B) 48
C) 84	D) 108

Answer & Explanation Answer: C) 84

Explanation:

Let the numbers be x and 4x. Then, $x \times 4 x = 84 \times 21 \Leftrightarrow x^{2} = \frac{84 \times 21}{4} \Leftrightarrow x = 21$

Hence Larger Number = 4x = 84

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Filed Under: HCF and LCM

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