**Q1. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:**

(a).173 m

(b).200 m

(c).273 m

(d).300 m

**Q2. A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has?**

(a).45

(b).60

(c).75

(d).90

**Solution 2(D)**

Let number of notes of each denomination be x.

Then x + 5x + 10x = 480

16x = 480

x = 30.

Hence, total number of notes = 3x = 90.

**Q3. David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross?**

(a).19

(b).28

(c).30

(d).37

**Solution 3(C)**

Suppose their paths cross after x minutes.

Then, 11 + 57x = 51 – 63x

120x = 40

x =1/3

Number of floors covered by David in (1/3) min. =1/3* 57= 19.

So, their paths cross at (11 +19) i.e., 30th floor.

**Q4. The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?**

(a).40

(b).50

(c).120

(d). 60

**Solution4(D)**

Let breadth = x metres

Then, length = (x + 20) metres.

Perimeter =5300/26.5 = 200 m.

2[(x + 20) + x] = 200

2x + 20 = 100

2x = 80

x = 40.

Hence, length = x + 20 = 60 m.

**Q5. The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is the two-digit number?**

(a) 69

(b) 78

(c) 96

(d) Cannot be determined

**Solution (d)**

Let the ten’s digit be x and unit’s digit be y.

Then, x + y = 15 and x – y = 3 or y – x = 3.

Solving x + y = 15 and x – y = 3, we get : x = 9, y = 6

Solving x + y = 15 and y – x = 3, we get : x = 6, y = 9

So the number is either 96 or 69. Hence, the number cannot be determined.

**Q6 Mr. Jones gave 40% of the money he had, to his wife. He also gave 20% of the remaining amount to each of his three sons. Half of the amount now left was spent on miscellaneous items and the remaining amount of Rs. 12,000 was deposited in the bank. How much money did Mr. Jones have initially?**

(a) 120000

(b) 110000

(c) 100000

(d) 150000

**Solution(c)**

Let the initial amount with Mr. Jones be Rs x. Then,

Money given to wife = Rs. (40/100) x= Rs. 2x/5. Balance = Rs. (x-2x/5) = Rs. 3x/5

Money given to 3 sons = Rs. [3×(20/100×3x/5)]=Rs.9x/25

Balance = Rs. (3x/5-9x/25)=Rs.6x/25

Amount deposited in bank = Rs. (1/2×6x/25)=Rs.3x/25

∴3x/25=12000⇔x=((12000 × 25)/3)=100000

**Q7. Milk contains 5% water. What quantity of pure milk should be added to 10 litres of milk to reduce this to 2%?**

(a) 5 litres

(b) 7 litres

(c) 15 litres

(d) Cannot be determined

**SOLUTION(C)**

Quantity of water in 10 litres = 5% of 10 litres = 0.5 litres

Let x litres of pure milk be added. Then, 0.5/(10 + x)=2/100⇔2x=30⇔x=15

Q**8. An article is sold at a certain price. By selling it at 2/3 of that price one loses 10%. The gain per cent at original price is :**

(a) 33 (1/3 )%

(b) 35%

(c) 40%

(d) 20%

**SOLUTION(b)**

New

S.P.=90

X=100

Gain

percent at original price=135-100/100=35%

**Q9. A vendor buys oranges at Rs. 2 for 3 oranges and sells them at a rupee each. To make a profit of Rs. 10, he must sell : **

(a) 10 oranges

(b) 20 oranges

(c) 30 oranges

(d) 40 oranges

**Solution 9(c)**

suppose he sells x oranges.Then,

C.P. = Rs.(2/3*x) = Rs.2x/3 and S.P. = Rs.x

so x -2x/3 = 10 or x = 30

**Q10. In how many different ways can 4 boys and 3 girls be arranged in a row such that all the boys stand together and all the girls stand together?**

(a) 75

(b) 576

(c) 288

(d) 24

**Solution 10(c)**

Required number of ways = 4! * 3! * 2! = 288.