**Q1. The price of an article in raised by 30% and then two successive discounts of 10% and 5% are allowed. What is the net increase in the price of the article**

(a) 16.66%

(b) 20%

(c) 11.11%

(d) 11.15%

**S1. Ans.(d)**

**Sol. **Selling price = 100 × 130/100 × 90/100 ×95/100

= 111.15

Net increase in price = 11.15%

**Q2. The population of the village increased by 10% in one year and in next year it is decrease by 10%. If at the end of second year. The population is 39600. What was the population at the beginning of first year.**

(a) 39800

(b) 40,000

(c) 44,000

(d) 42,000

**S2. Ans.(b)**

**Sol.** Population at the beginning of a first year

= 39600 ×100/(100+10)×100/(100-10)

= 40,000

**Q3. A man’s income is increased by Rs. 2000 in a year rate of income tax having reduced from 25% to 15%, his tax liability remains the same as in the least year. Find his income in the current year.**

(a) 2000

(b) 5000

(c) 6000

(d) 3000

**S3. Ans.(b)**

**Sol**. Decrease in tax rate = 25% – 15% = 10%

Relative decrease in tax rate = 10/25=2/5

If the liability is same,

increase in income = 2/(5-2) of last year income = 2000

Income during last year = 3000

Income during current year = 3000 + 2000 = 5000

**Q4. If Price of sugar having been fallen by 10%. then consumer can buy 21 kg. more than before. Had the price been increased by 16.6% how much quantity of sugar could he have bought for the same sum.**

(a) 180 kg

(b) 162kg

(c) 210 kg

(d) None of these

**S4. Ans.(b)**

**Sol. **Decrease in quantity = 10% = 1/10

Quantity brought is increased by 1/9 of original quantity

Original quantity = (9 × 21) kg

= 189 kg.

If price is increased by 16.66% = 1/6

So, decreased quantity = 1-1/7=6/7

Quantity brought = 9 × 21 × 6/7

= 162 kg.

**Q5. What is the remainder when 1044 × 1047 × 1050 × 1053 is divided by 33?**

(a) 13

(b) 22

(c) 30

(d) 8

**S5. Ans.(c)**

**Sol.** Remainder = (1044 × 1047 × 1050 × 1053)/33

=(21×24×27×30)/33

=(21×24×-6×-3)/33( by using remainder theorem)

=(21×24×18)/33=(-12×-9×18)/33

⇒(12×9×18)/33==30/33

Remainder = 30

**Q6. Rs. 4000 is divided into two parts such that one part is put out at 3% and the other at 5% rate. If the annual interest earned from both the investments be Rs. 144, find the first part.**

(a) Rs. 4000

(b) Rs. 2800

(c) Rs. 3800

(d) Rs. 3500

**S6. Ans.(b)**

**Sol.** Average rate = 144/4000 × 100 = 3.6%

Ratio = 14:6 = 7:3(using alligation method

So, first part = 7/10 × 4000 = Rs. 2800

**Q7. The average age of 24 students and the class teacher is 16 years. If the class teacher’s age is excluded, the average is reduced by one year. Find the age of the class teacher.**

(a) 40 years

(b) 38 years

(c) 50 years

(d) 56 years

**S7. Ans.(a)**

**Sol. **Class teacher’s age = 25 × 16 – 24 × 15

= 400 – 360 = 40 years

**Q8. A person saves 15% of his income. If his income increase by 35% and he starts savings 20% of his income, by what percentage does his savings increase?**

(a) 60%

(b) 70%

(c) 75%

(d) 80%

**S8. Ans.(d)**

**Sol.** Assume his initial income = Rs. 100, so savings = 15

New income = 135 ⇒ new savings = 27

Percentage increase in savings = 80%

**Q9. An amount doubles itself in 5 years at S.I. In how many years will its interest become its 250%?**

(a) 10 years

(b) 15/2 years

(c) 25/2 years

(d) 15 years

**S9. Ans.(c) **

**Sol.** Amount doubles itself → 100% is added to the principal in five years → 20% is added to principal every year.

So, Rate of interest = 20%

For interest to become 250%, time required = 250/20 = 12.5 years

**Q10. D, K and A can do a work in 5, 15 and 35 days respectively. They get an amount of Rs. 1054 for finishing the work working together. What is the share of K in that amount?**

(a) Rs. 242

(b) Rs. 324

(c) Rs. 238

(d) Rs. 245

**S10. Ans.(c)**

**Sol.** D share: K share : A share

= K time × A time : D time × A time : D time × K time

= 15 × 35 : 5 × 35 : 5 × 15 = 21 : 7 : 3

So, K’s share = 1054/31 × 7 = Rs. 238

**Q11. There are two alloys made up of copper and gold. In the 1st alloy, copper is half of the gold and in the second alloy copper is thrice as much as gold. How many times the second alloy must be mixed with first alloy to get the new alloy in which copper is twice as that of gold?**

(a) 3

(b) 4

(c) 5

(d) 6

**Q12. If x = b + c – 2a, y = c + a – 2b, z = a + b – 2c, then the value of x^2+y^2-z^2+2xy is**

(a) 0

(b) a + b + c

(c) a – b + c

(d) a + b – c

**Q13. The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers?**

(a) 21

(b) 25

(c) 41

(d) 67

**S13. Ans.(c)**

**Sol**. Using options, we find that four consecutive odd numbers are 37, 39, 41 and 43. The sum of these 4 numbers is 160, when divided by 10, we get 16 which is a perfect square.

Therefore, 41 is one of the odd numbers.

**Q14. When you reverse the digits of the number 13, the number increases by 18. How many other two digit numbers increase by 18 when their digits are reversed?**

(a) 5

(b) 6

(c) 7

(d) 8

**S14. Ans.(b)**

**Sol**.Let the number be (10x+y), so when the digits of number are reversed the number becomes (10y+x). According the question, (10y+x)-(10x+y)=18

9(y-x)=18 y-x=12

So, the possible pairs of (x, y) are

(1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9).

But we want the number other than 13. Thus, there are 6 possible numbers ie, 24, 35, 46, 57, 68, 79.

So, total number of possible numbers are 6.

**Q15. The number of employees in Obelix Menhir Co. is a prime number and is less than 300. The ratio of the number of employees who are graduates and above, to that of employees who are not, can possibly be**

(a) 101 : 88

(b) 87 : 100

(c) 110 : 111

(d)97 : 84

**S15. Ans.(d)**

**Sol. **Using options, we find that sum of numerator and denominator of 97 : 84 is (97+84)=181 which is a prime number. Hence, it is the appropriate answer.