**Q1. Sum of cost price of two watches is Rs 3360. One watch is sold at a profit of 12% and other at a loss of 12%. Due to this transaction, there is neither profit nor loss. Find the cost price of each watch.**

**(a) Rs 1680, Rs 1680**

(b) Rs 1280, Rs 1280

(c) Rs 1000, Rs 1000

(d) Rs 2100, Rs 2160

Sol.

On one watch, he earns 12% profit and on the other one 12% loss. There is, finally, no profit no loss.

Hence CPs of both the watches should be same such that their summation = Rs 3360. Hence CP/watch = (Rs.3360)/2 = Rs. 1680

Hence option (a) is the answer.

**Q2. A seller buys 50 kg oranges for Rs 525 and gains a profit equal to the cost price of 10 kg of oranges. Find the selling price per kg of oranges (Rs/kg).**

**(a) 12.60**

(b) 11.60

(c) 10.20

(d) 12

(e) 13.125

Sol.

Assume SP/kg = Rs x/kg.

CP/kg =(Rs.525)/50 = Rs. 10.5/kg

Total SP – Total CP = profit ⇒ 50x – Rs. 525 = 10 × Rs. 10.5 = Rs. 105

Or, 50x = Rs. 105 + Rs. 525 = Rs. 630

Hence x =(Rs.630)/50 = Rs. 12.6/kg

**Q3. Find a number, one-seventh of which exceeds its eleventh part by 100.**

**(a) 1925**

(b) 1825

(c) 1540

(d) 1340

Sol.

Let the number be x.

x/7-x/11 = 100

=>(11x-7x)/(11×7)=100

=> 4x = 77 x 100

=> x = 1925

**Q4. 6 bells commence tolling together and toll of intervals are 2, 4, 6, 8, 10 and 12 s, respectively. In one hour, how many times do they toll together?**

(a) 16

(b) 32

(c) 21

**(d) 31**

Sol.

LCM of 2, 4, 6, 8, 10, 12 = (2 × 2 × 3 × 2 × 5) = 120

∴ After every 2 min they toll together

∴ Number of times they toll together in one hour

=(60/2+1) times = 31 times

**Q5. A certain type of board is sold in lengths of multiples of 2 ft. the shortest board sold is 6 ft and the longest is 24 ft. A builder needs a large quantity of these types of board in 5 1/2 ft lengths. Find the length of be ordered for which waste is minimum.**

(a) 24 ft

(b) 26 ft

**(c) 22 ft**

(d) 52 ft

Sol.

LCM of (2 and 5 1/2) = LCM of (2 and 11/2)

Required answer =(LCM of 2 and 11)/(HCF of 1 and 2)=22/1 = 22 ft

**Q6. The average expenditure of a man for the first 12 days of a month was Rs. 850 per day and for the next 18 days, it was Rs. 825 per day. If he could save Rs. 700 during that month. Find his income for the month.**

(a) Rs. 22650

(b) Rs. 29450

(c) Rs. 24150

**(d) Rs. 25750**

Sol.

Expenditure for the first 12 days (850 × 12) = Rs. 10200

Expenditure for the next 18 days = (825 × 18) = Rs. 14850

Savings = Rs. 700

∴ Income for the month = (10200 + 14850 + 700) = Rs. 25750

**7. If each side of a rectangle is increased by 50%, its area will increase by**

**(a) 125%**

(b) 75%

(c) 100%

(d) 225%

Sol.

Original area = l × b

New area =3/2 l×3/2 b=9/4 lb

Increase % =(9/4 lb – lb)/lb×100%

= 125%

**8. If x^2+1/x^2 =34, then find the value of x+1/x**

(a) 4

**(b) 6**

(c) 8

(d) 17

Sol.

x^2+1/x^2 =34

(x+1/x)^2-2×1/x=34

[a^2+b^2=(a+b)^2-2ab ]

(x+1/x)^2=36

x+1/x=6

**9. The value of 5-(2 1/2-3/4)+(3 1/2-1 1/4) is**

(a) 4 1/2

**(b) 5 1/2**

(c) 5 1/4

(d) 3 1/2

Ans. (b)

Sol.

5-[5/2-3/4]+[7/2-5/4]=5-[(10 – 3)/4]+[(14-5)/4]=5-7/4+9/4

=(20-7+9)/4=22/4=11/2=5 1/2

**10. One-fourth of a pizza was eaten by Renu. The rest was equally distributed among 12 children. What part of the pizza did each of these children get?**

**(a) 1/16**

(b) 1/32

(c) 3/16

(d) 1/8

Sol.

Pizza remaining = 3/4

Pizza that single child get = (3/4) × (1/12)=1/16