**Q1. The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is: **

(a) 15

(b) 16

(c) 18

(d) 25

**1. Ans.(a)**

**Sol. **

Selling price of x article = cost price of 20 article

Selling price of x article – cost price of x article

= cost price of (20 – x) article

Profit = cost price of (20 – x) article

% profit =(Cost price of (20-x) article )/(Cost of 20 article)

25%=(20 – x)/20

=1/4=(20 – x)/20

20 = 80 – 4x

4x = 60

x = 15

**Q2. A shopkeeper expects a gain of 22.5% on his cost price. If in a week, his sale was of Rs. 392, what was his profit? **

(a) Rs. 18.20

(b) Rs. 70

(c) Rs. 72

(d) Rs. 88.25

**2. Ans.(c)**

**Sol.** Let the cost price = 100x

Sell price = 122.5x

According to question,

122.5x = 392

x=392/122.5=3.2

100x = 320

Profit = 392 – 320 = 72

**Q3. On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:**

(a) Rs. 45

(b) Rs. 50

(c) Rs. 55

(d) Rs. 60

**3. Ans.(d)**

**Sol.** Selling of 17 ball = cost price of 17 – cost price of 5 ball

Selling of 17 ball = cost price of 12 ball

Cost price of 12 ball = 720

Cost price of ball =720/12=60

**Q4. A series of discounts of 30%, 20% and 10% is equivalent to a single discount of ….**

(a) 70%

(b) 49.6%

(c) 48.4%

(d) 49.4%

**4. Ans.(b)**

**Sol.** x is a mark price

Selling price =x(1-30/100)(1-20/100)(1-10/100)

= x×.7×.8×.9

= .504x

Single discount % =((x – .504x))/x×100

⇒ 49.6%

**Q5. Mohan Sells 25 toys and gains the S.P. of 9 toys. Find his percentage profit.**

(a) 56.2%

(b) 60%

(c) 40%

(d) 56.4%

**5. Ans.(a)**

**Sol. **Cost price of 25 toys = sell price of 16 toys

(Sell price)/(Cost price )=25/16

% profit =(SP/CP-1)×100

=9/16×100

= 56.2%

**Q6. If two numbers are in the ratio 6 :13 and their least common multiple is 312, then what is the sum of numbers?**

(a) 57

(b) 67

(c) 75

(d) 76

**6. Ans.(d)**

**Sol.**

Let the first number = 6x

Second number = 13x

LCM of first and second number = 6 × 13 × x

6 × 13x = 312

x=312/(6×13)

x = 4

Sum of number = 6 × 4 + 13 × 4 = 76

**Q7. When n is divided by 4, the remainder is 3. What is the remainder when 2n is divided by 4?**

(a) 1

(b) 3

(c) 2

(d) None of these

**7. Ans.(c)**

**Sol. **

n = 4x + 3

2n = 2(4x + 3)

2n = 8x + 6

According to question,

Remainder =(8x + 6)/4

remainder = 2

**Q8. Find the smallest number required to be added to 2000 to get a number exactly divisible by 17.**

(a) 6

(b) 4

(c) 2

(d) 3

**8. Ans.(a)**

**Sol. **

When 2000 is divided by 17 remainder is equals to 11

Smallest number required to be added = 17 – 11 = 6

**Q9. Two numbers are in the ratio 3: 4, and their LCM is 108. Find their HCF.**

(a) 9

(b) 8

(c) 6

(d) 5

**9. Ans.(a)**

**Sol. **

Let the first number = 3x

Second number = 4x

LCM of these two number = 3 × 4x = 12x

12x = 108

x = 9

HCF of these two number = x

= 9

**Q10. A number P4571203K is divisible by 18. Which of the following values can P and K take?**

(a) 1, 2

(b) 2, 3

(c) 3, 3

(d) 6, 8

**10. Ans.(d)**

**Sol. **

Number = P4571203K

If the number is divisible by 18, then it should be divisible by 2 and 9

So, P + 4 + 5 + 7 + 1 + 2 + 0 + 3 + k = 22 + P + K

⇒ P + K = 5 or P + K = 14

And K must be even number or zero

So, from the following option (d) option fulfilled the condition.

**Q11. Six-elevenths of a number is equal to 22% of second number. Second number is equal to the one-fourth of third number. The value of the third number is 2400. What is the 45% of the first number?**

(a) 109.8

(b) 111.7

(c) 117.6.0

(d) None of these

**11. Ans.(d)**

**Sol. **

Third number = 2400

Second number =1/4×2400=600

First number ×6/11=22% of 600

First number =11/6×600×22/100=242

45% of the first number =(45 × 242)/100=108.90

**Q12. In an entrance examination Ritu scored 56% marks, Smita scored 92% marks and Rina scored 634 marks. The maximum marks of the examination are 875, what are the average mark scored by all the three girls together:**

(a) 929

(b) 815

(c) 690

(d) 643

**12. Ans.(d)**

**Sol. **

Maximum marks in examination = 875

Ritu’s marks =875×56/100=490

And Smita’s marks =875×92/100=805

Hence, required average marks =(490 +805 + 634)/3

= 643

**Q13. If the denominator of a fraction is increased by 350%, and numerator is increased by 150%, then the resultant fraction is 25/51, what is the original fraction:**

**13. Ans.(c)**

**Sol.**

**Q14.In An election there were only two candidates, one of the candidates secured 40% of votes and is defeated by the other candidate by 298 votes. The total number of votes polled is:**

(a) 745

(b) 1460

(c) 1490

(d) 1500

**14. Ans.(c)**

**Sol. **

Let the total number of votes polled be x%

Now, according to the question

x×((60-40)/100)=298

x=298×5=1490

**Q15. In a test consisting of 80 questions carrying one mark each, Arpita answer 65% of the first 40 question correctly, what percent of the other 40 question does she need to answer correctly to score 75% on the entire test:**

(a) 60

(b) 85

(c) 75

(d) 40

**15. Ans.(b)**

**Sol.**

⇒(60-40 × 65/100 )/40× 100=34/40×100=85%