**1. Matt, Maria and Mona went to a party. They played a total of 18 games of poker. Each game costs Rs. 100 to play. On winning a game, winner gets a prize of Rs. 500. If only the games played by Matt and Mona are considered, there is a profit of Rs. 100, and if only the games played by Matt and Maria are considered, there is a loss of Rs. 1500. How many games did Matt play?**

b) 1

c) 6

d) 11

e) Cannot be determined

**2. In a rhombus, the length of a diagonal is 49 cm less than the other diagonal. The area of the rhombus is 1911 square cm. Find the side length of the rhombus.**

a) 43.13 cm

b) 46.54 cm

c) 49.42 cm

d) 50.11 cm

e) 53.46 cm

**3. A certain sum of money is invested at compound interest for 2 years at 20 per cent per annum. If the interest is added half-yearly instead of per annum, then the interest increases by Rs. 482, what is the sum of money?**

a) Rs. 50000

b) Rs. 30000

c) Rs. 20000

d) Rs. 10000

e) Rs. 12000

**4. P has two sons, Q and R. P bought 80 kg of rice at Rs. 40 per kg. He gave some amount to Q and remaining amount to R, and asked them to sell the rice. Q sold the rice at Rs. 44 per kg. R sold the rice at Rs. 40 per kg but using a 1 kg weight that actually weighs 900 grams only. In total, they together earned a profit of 10.5%. How much rice was given to R?**

a) 32 kg

b) 36 kg

c) 40 kg

d) 48 kg

e) 44 kg

**5. A five digit even number is to be formed by using digits 3, 5, 7, 8 and 9. Each digit can be used at most twice. How many such numbers can be formed?**

a) 204

b) 360

c) 240

d) 444

e) 480

**6. Train P and Train Q, both have a length of 500 metres. Q is stationary and P is running at a speed of 72 km/hr on a parallel track. When P is crossing Q, the moment at which the heads of both trains are at same point, train Q also start running at 36 km/hr. How much time in total will train P take to cross train Q (in seconds)?**

a) 75

b) 90

c) 105

d) 120

e) 135

**7. Matt and Mark can finish a work alone in 40 days and 20 days, respectively. They work together and finish half of the work. After that, Matt stops working for 5 days. Then, he joins back Mark in work. How many more days will they take to finish the work?**

a) 5

b) 10/3

c) 4

d) 8/3

e) 11/3

**8. There are three different varieties of vinegar. First has 80% acid and rest is water. Second has 60% acid and rest is water and third has 50% acid and rest is water. If the three are mixed in ratio 2 : 3 : 4, what will be the percentage of water in the mixture?**

a) 30

b) 40

c) 45

d) 50

e) 60

**9. There are 420 students residing in a hostel. If the students increase by 100, the expenses of the mess increase by Rs 500 per month, while the average expenditure per head diminishes by Rs 25. What is the original monthly expense of the mess?**

a) Rs 105

b) Rs 120

c) Rs 135

d) Rs 180

e) Rs 165

**10. A sum of Rs 6800 is deposited in two schemes. One part is deposited in Scheme A which offers 12% rate of interest. Remaining part is invested in Scheme B which offers 10% rate of interest compounded annually. If interest obtained in scheme A after 5 years is Rs 678 more than the interest obtained in scheme B after 2 years, find the part deposited in scheme A.**

a) Rs 4200

b) Rs 3900

c) Rs 3400

d) Rs 2600

e) Rs 3000

**Solutions:**

**1. E)**Suppose Matt, Maria and Mona played P, Q and R games, respectively.

⇒ P + Q + R =
18

If only the
games played by Matt and Maria are considered, there is a loss of Rs. 1500. So,
Matt and Maria have played at least 15 games between them.

If they have
played 15 games, they have lost all of them. If they have played 16 games, they
can lose Rs. 1600 by losing all games or lose Rs. 1000 by losing 15 games and
winning 1, or lose less than that by winning more games. So, this case is not
possible as they cannot lose Rs. 1500. Similarly, if they have played 17 games,
they can lose Rs. 1700 by losing all games or lose Rs. 1100 by losing 16 games
and winning 1, or lose less than that by winning more games. And, if they have
played 18 games, they can lose Rs. 1800 by losing all games or lose Rs. 1200 by
losing 17 games and winning 1, or lose less than that by winning more games.

⇒ Only case
possible is that they have played 15 games and lost all of them.

⇒ P + Q = 15

⇒ R = 18 – 15 = 3

If only the
games played by Matt and Mona are considered, there is a profit of Rs. 100. We
know that Matt has lost all of his P games. Mona could have won 1, 2 or 3
games.

If Mona had
won 1 game, total profit = 500 – 100 × (P + 3) = 100

⇒ P + 3 = 4

⇒ P = 1

If Mona had
won 2 games, total profit = 2 × 500 – 100 × (P + 3) = 100

⇒ P + 3 = 9

⇒ P = 6

If Mona had
won 3 games, total profit = 3 × 500 – 100 × (P + 3) = 100

⇒ P + 3 = 14

⇒ P = 11

∴ Matt could
have played 1, 6 or 11 games. Answer cannot be uniquely determined.

**2. D)**Let the length of a diagonal be T cm and another diagonal be (T - 49) cm.

The area of
the rhombus is 1911 square cm.

⇒ (1/2) × T × (T – 49) = 1911

⇒ T

^{2}– 49T – 3822 = 0
⇒ (T + 42) (T
- 91) = 0

Since diagonal
length cannot be negative, T = 91.

⇒ Diagonals
have lengths 91 cm and 42 cm.

In a rhombus,
diagonals bisect each other perpendicularly.

∴ Side length
= √ [(91/2)

^{2}+ (42/2)^{2}] = 50.11 cm**3. C)**Let the principal be Rs. x

Rate of
interest = 20% p.a.

Time period,
t = 2 years

We know the
formula for compound interest – CI = [P { (1 + r/100)

^{t}– 1 } ]
Where, CI =
Compound interest, P = Principal, R = Rate of interest, T = Time period

Thus,
according to question

Now, according
to condition given in the question

4641x/10000 =
11x/25 + 482

4641x/10000 =
(11x + 12050)/25

4641x/400 =
(11x + 12050)

4641x =
400(11x + 12050)

4641x = 4400x
+ 4820000

241x =
4820000

∴ x = 20000

Thus, the
principal will be Rs. 20,000

**4. B)**Suppose P gave T kg of rice to R and (80 – T) kg of rice to Q.

Total cost
price of rice = Rs. 40 × 80 = Rs. 3200

Selling price
of rice sold by Q = Rs. 44 × (80 – T) = Rs. (3520 – 44T)

R sold the
rice at Rs. 40 per kg but using a 1 kg weight that actually weighs 900 grams
(or 0.9 kg) only.

So, with T kg
rice, R can take selling price of (T/0.9) kg of rice.

⇒ Selling
price of rice sold by R = Rs. 40 × (T/0.9) =
Rs. 400T/9

Total selling
price = Rs. (3520 – 44T + 400T/9)

Profit earned
is 10.5%.

We know,
Selling price = Cost Price × (1 + (Profit Percentage)/100)

⇒ (3520 – 44T + 400T/9) = 3200 × (1 + 10.5/100) = 3536

⇒ 4T/9 = 16

⇒ T = 36

∴ 36 kg of
rice was given to R.

**5. D)**If the number is even, its last digit must be 8, as all other digits are odd.

For remaining
4 digits, we can have two different cases.

**Case i) 8 is one among the first four digits.**

Here, 8 can
take any of the four positions. For remaining 3 positions, all three digits can
be different or two digits are same and one is different.

If all three
digits are different, it can be done in

^{4}C_{3}ways, i.e., 4 ways. After that they can be placed in 6 different ways.
If two digits
are same and one is different, then those two digits can be chosen in 2 ×

^{4}C_{2}ways, i.e., 12 ways. After that they can be placed in 3 different ways.
So, total
numbers that can be formed in this case = 4 × (4 × 6 + 12 × 3) = 240

**Case ii) 8 is not one among the first four digits.**

For remaining
4 positions, all four digits can be different or two digits are same and two
are different or there are two pairs of two different digits.

If all four
digits are different, it can be done in

^{4}C_{4}ways, i.e., 1 way. After that they can be placed in 24 different ways.
If two digits
are same and two are different, then those three digits can be chosen in 3 ×

^{4}C_{3}ways, i.e., 12 ways. After that they can be placed in 4!/2! = 12 different ways.
If there are
two pairs of two different digits, then those two digits can be chosen in

^{4}C_{2}ways, i.e., 6 ways. After that they can be placed in 4!/2!2! = 6 different ways.
So, total
numbers that can be formed in this case = (1 × 24 + 12 × 12 + 6 × 6) = 204

∴ Total
possible numbers = 240 + 204 = 444

**6. A)**Speed of train P = 72 km/hr = 72 × (5/18) m/sec = 20 m/sec

To reach from
the rear of train Q to front of train Q, train P will have to travel a distance
equal to length of train Q. (Train Q is stationary in this period)

Time taken =
Length of train Q/Speed of train P = 500/20 = 25 seconds

To reach from
the front of train Q to completely cross train Q, train P will have to travel a
distance equal to its own length. (Train Q is running in this period)

Speed of
train Q = 36 km/hr = 36 × (5/18) m/sec = 10 m/sec

Relative
speed of train P with respect to train Q = 20 m/sec – 10 m/sec = 10 m/sec

Time taken =
Length of train P/Relative speed of train P with respect to train Q = 500/10 =
50 seconds

∴ Total time
taken = 25 seconds + 50 seconds = 75 seconds

**7. B)**Matt and Mark can finish a work alone in 40 days and 20 days, respectively.

⇒ In a day,
Matt and Mark finish 1/40 and 1/20 of work, respectively.

They work
together and finish half of the work. After that, Matt stops working for 5
days. Then, he joins back Mark in work. Suppose they take N more days to finish
the work.

⇒ ½ + 5(1/20) + N(1/40 + 1/20) = 1

⇒ N (3/40) = ¼

⇒ N = 10/3

∴They took
10/3 more days to finish the work.

**8. B)**Let the quantities of three varieties mixed be 2T, 3T and 4T, respectively.

Percentage of
acid in resultant mixture = [(2T × Percentage of acid in first variety) + (3T ×
Percentage of acid in second variety) + (4T × Percentage of acid in third
variety)] / (2T + 3T + 4T)

⇒ [(2T × 80) + (3T × 60) + (4T × 50)]/9T = 540T/9T = 60

∴ Percentage
of water in resultant mixture = 100 - Percentage of acid in resultant mixture =
100 – 60 = 40%

**9. C)**Let original monthly expense of the mess = Rs x

So 520 (x –
25) – 420x = 500 [520 is (420+100)]
Solve, x = Rs 135

**10. D)**(6800-x)*12*5/100 = x [ (1 + 10/100)

^{2}– 1] + 678

68*12*5 –
60x/100 = 21x/100 + 678

Solve, x = Rs
4200

So in scheme
A = 6800-4200 = Rs 2600