**Q1. The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:**

(a) 298

(b) 286

(c) 330

(d) 308

** S1. Ans.(d)**

**Sol**. Let the other number be x. Then,

11 × 7700 = 275 × x

⇒x=(11×7700)/275=308

Hence, the other number is 308.

**Q3. What is the greatest number that divides 263, 935 and 1383 leaving a remainder of 7 in each case?**

(a) 30

(b) 31

(c) 32

(d) 34

**S3. Ans.(c)**

**Sol.** The required greatest number is the HCF (263 – 7, 935 – 7, 1383 – 7)

i.e., 256, 928 and 1376 HCF = 32

**Q4. What is the sum of the digits of the least number which when divided by 54 leaves 35 as remainder, when divided by 80 leaves 61 as remainder and when divided by 119 leaves 100 as remainder?**

(a) 21

(b) 19

(c) 25

(d) 28

**S4. Ans.(d)**

**Sol.** (54 – 35) = 19, (80 – 61)

= 19 and (119 – 100) = 19

LCM of 54, 80 and 119 is 257040.

Required least number = 257040 + 19 = 257059

Sum of the digits of 257059 = 2 + 5 + 7 + 0 + 5 + 9 = 28

**Q6. In a family of 5 members, the average age at present is 33 years. They youngest member is 9 years old. The average age of the family just before the birth of the youngest member was**

(a) 30 years

(b) 29 years

(c) 25 years

(d) 24 years

**S6. Ans.(a)**

**Sol. **Sum of the present age of family members

= 33 × 5 = 165 years

9 years ago,

Sum of their ages = 165 – 9 × 5 = 120 years

∴ Required average age = 120/4 = 30 years

**Q7. The mean of 100 items was 46. Later on it was discovered that an item 16 was misread as 61 and another item 43 was misread as 34. It was also found that the number of items was 90 and not 100. Then what is the correct mean?**

(a) 50

(b) 50.7

(c) 52

(d) 52.7

**S7. Ans.(b)**

**Sol.** Required Average

=(100×46-61-34+16+43)/90

=(4600-36)/90=4564/90=50.7

**Q8. In an examination, 10% of the students failed in Physics, 20% failed in Chemistry and 5% failed in both. What is the percentage of students who failed in atleast one subject?**

(a) 70%

(b) 75%

(c) 80%

(d) 65%

**Q9. In an examination, it is required to get 296 marks out of total marks to pass. A student gets 222 marks and is declared failed by 10% marks. What are the maximum total marks as student can get?**

(a) 830

(b) 810

(c) 780

(d) 740

**S9. Ans.(d) **

**Sol.**Let the maximum aggregate marks be x.

10% of x = 296 – 222

**Q10. Anupam purchased 5 Lemons for Rs. 4. He sold all the Lemons and earned 75% Profit. At what rate did he sell the each Lemons?**

(a) Rs. 2.4

(b) Rs. 1.4

(c) Rs. 1.8

(d) Rs. 2

**S10. Ans.(b)**

**Sol.**Directly: Cost price for 5 Lemon is Rs. 4

So SP for 5 Lemon = 4 × ((100 + 75))/100 = Rs. 7

So, SP for 1 Lemon = 7/5 = Rs. 1.4

**Q11. A person sells a T-shirt at a profit of 10%. If he had bought it at 10% less and sold it for Rs. 3 more, he would have gained 25%. Find the cost price.**

(a) Rs. 120

(b) Rs. 150

(c) Rs. 240

(d) Rs. 180

**Q12. A sells a bicycle to B at a profit of 30% and B sells it to C at a loss of 20%. If C pays Rs. 520 for it, at what price did A buy?**

(a) Rs. 480

(b) Rs. 450

(c) Rs. 400

(d) Rs. 500

**S12. Ans.(d)**

**Sol. **In the whole transaction there is a gain of 30% and a loss of 20% so our calculating figures would be 130, 80 on 100

B’s cost of price = 520 × 100/80

A’s cost price = 520 × 100/80×100/130 = Rs. 500

**Q13. Mr. x and Mr. y each bought the same motorcycle using a 10% off coupon. Mr. x’s cashier took 10% off the price and then added 8.5% sales tax whereas Mr. y’s cashier first added the sales tax and then took 10% off the total price. The amount Mr. x paid is**

(a) same as the amount Mr. y paid

(b) greater than Rs. 850 as the amount Mr. y paid

(c) less than Rs. 550 as the amount Mr. y paid

(d) greater than Rs. 85 as the amount Mr. y paid

**S13. Ans.(a)**

**Sol. **Amount paid by Mr. x= 100×90/100×108.5/100

Amount paid by Mr. y =100×108.5/100×90/100

Hence, option(a) is correct

**Q14. Rs. 1,000 is invested at 5% per annum simple interest. If the interest is added to the principal after every 10 years, the amount will become Rs. 2,000 after **

(a) 15 years

(b) 18 years

(c) 20 years

(d) 50/3 years

**S14. Ans.(d)**

**Sol. **SI(10)=(1000×5×10)/100

=500

Now principal = 1500

500=(1500×S×T)/100

6 2/3=T

Total time =10+20/3

=50/3

**Q15. A sum of money at some rate of simple interest amounts to Rs. 2900 in 8 years and to 3,000 in 10 years. The rate of interest per annum is **

(a) 4%

(b) 5/2%

(c) 3%

(d) 2%

**S15. Ans.(d)**

**Sol. **Let the Principal = P

(2900-P)=(P×r×8)/100 ………….(i)

(3000-P)=(P×r×10)/100 ………..(ii)

Divided (i) by (ii)

29000-10P=24000-8P

5000=2P

P=2500

400=(2500×8×r)/100

r=2%