1. The value of (0.03)2 – (0.01)2 / 0.03 – 0.01 is

(1) 0.02

(2) 0.004

(3) 0.4

(4) 0.04

### Show Answer

Answer : (4) 0.004

**Explanation :**

**(4) (0.03)2 -(0.01)2 / 0.03 – 0.01**

**[Using a2 – b2 = (a + b) (a – b)]**

**= (0.03 + 0.01)(0.03 – 0.01) / 0.03 – 0.01**

**= 0.03 + 0.01 = 0.04**

2. The H.C.F of two numbers is 8. Which one of the following can never their LC.M?

(1) 24

(2) 48

(3) 56

(4) 60

### Show Answer

Answer : (4) 60

**Explanation :**

**(4) HCF of two numbers is 8. This means 8 is a factor common to both the numbers. LCM is common multiple for the two numbers. It is divisible by the two numbers. So, the requird answer = 60**

3. A B and C rent a pasture. A puts in 10 oxen for 7 months. B 12 oxen for 5 months and C 15 oxen fro 3 months for grazing. If the rent of the pasture is Rs. 175 /. how much must C pay as his share of rent?

(1) Rs 45/-

(2) Rs 50/-

(3) Rs 55/-

(4) Rs 60/-

### Show Answer

Answer : (1) Rs 45/-

**Explanation :**

**(1) Share of rent = (number of oxen x time)**

**A : B : C**

**= (10 * 7) : (12 * 5) : (15 * 30)**

**A : B : C = 70 : 60 : 45**

**A : B : C = 14 : 12 : 9**

**C’s share of rent = 9 / 14 + 12 + 9 * 175**

**= 9 / 35 * 175 = 45**

**Therefore C’s share of rent is Rs. 45**

4. A reduction of 20% in the price of oranges enables a man to buy 5 oranges more for Rs. 10 / -. The price of an orange before reduction was :

(1) 20 paise

(2) 40 paise

(3) 50 paise

(4) 60 paise

### Show Answer

Answer : (3) 50 paise

**Explanation :**

**(3) 20% of Rs. 10**

**= 20 / 100 * Rs. 10 = Rs. 2**

**Reduced price of 5 oranges = Rs. 2**

**Therefore Reduced price of 1 orange**

**= Rs. 2 / 5 = 200 / 5 paise = 40 paise**

**Original price of 1 orange **

**= 40 / 1 – 0.20 = 40 / 0.8 = 400 / 8 = 50 paise**

5. In an examination, a student who gets 20% of the maximum marks fails by 5 marks. Another student who scores 30% of the maximum marks gets 20 marks more than the pass marks. The necessary percenatge required for passing is:

(1) 32%

(2) 23%

(3) 22%

(4) 20%

### Show Answer

Answer : (3) 22%

**Explanation :**

**(3) Let the maximum marks be x.**

**According to question.**

**20% of x + 5 = 30% of **

**x – 20**

**= (30 – 20)% of x = 25**

**= x = 25 * 100 / 10 = 250**

**Therefore Pass marks**

**= 20% of 250 + 5 = 55**

**Therefore percentage Pass marks**

**= 55 / 250 * 100 = 22%**

6. When 60 is subtracted from 60% of a number, the result is 60. The number is :

(1) 120

(2) 150

(3) 180

(4) 200

### Show Answer

Answer : (2) 150

**Explanation :**

**(2) Let the number be x. Then x – 60% of x = 60**

**= x – 0.60x = 60 = 0.4x = 60**

**= x = 60 / 0.4 = x = 600 / 4**

**x = 150**

**Therfore The number is 150.**

7. If 3 men of 6 women can do a piece of work in 16 daays, in how many days can 12 men and 8 women do the same piece of work?

(1) 4 days

(2) 5 days

(3) 3 days

(4) 2 days

### Show Answer

Answer : (3) 3 days

**Explanation :**

**(3) 3m = gw**

**Therefore m = 2w**

**12m + 8w = (12 * 2w) + 8w = 32w**

**Therefore 6 women can do the work in 16 days.**

**Therefore 32 women can do the work in**

**16 * 6 / 32 = 3 days**

8. On what sum does the difference between the compound interest and the simple interest for 3 years at 10% id Rs. 31?

(1) Rs. 1500

(2) Rs. 1200

(3) Rs. 1100

(4) Rs. 1000

### Show Answer

Answer : (4) Rs. 1000

**Explanation :**

**(4) Let the sum be Rs. x**

**r = 10%, n = 3 years**

**S.I. = x * r * n / 100**

**S.I = x * 10 * 3 / 100 = 3 / 10 x**

**C.I. = [(1 + R / 100)N – 1]x**

**= [(1 + 10 / 100)3 – 1]x = [(11 / 10)3 – 1]x**

**= (1331 / 1000 – 1)x = 331 / 1000 x**

**331 / 1000 x – 3 / 10x = 31**

**or, (331 – 300) / 1000 x = 31**

**or, 31 / 1000 x = 31 or, x = 1000**

**Therefore Sum = Rs. 1000**

9. At what percent per annum will Rs. 3000 amount to Rs. 3993 in 3 years if the interest is compounded annually?

(1) 9%

(2) 10%

(3) 11%

(4) 13%

### Show Answer

Answer : (2) 10%

**Explanation :**

**(2) P = Rs. 3000, A = Rs. 3993, **

**n = 3 years**

**A = P [1 + r / 100]n**

**Therefore [1 + r / 100]n = A / P**

**[1 +r / 100]3 = 3993 / 3000 = 1331 / 1000**

**[1 + r / 100]3 = [11 / 10]3**

**= 1 + r / 100 = 11 / 10 = r / 100 = 11 / 10 – 1**

**= r / 100 = 1 / 10 = r = 100 / 10**

**r = 10%**

10. Given that 0.111 …. = 1 / 9; 0.444

is equal to :

(1) 1 / 90

(2) 2 / 45

(3) 1 / 99

(4) 4 / 9

### Show Answer

Answer : (4) 4 / 9

**Explanation :**

**(4) Given : 0.111…. = 1 / 9**

**0.444….. = 4 * 0.1111….**

**= 4 * 1 / 9 = 4 / 9**