**Q1.A merchant purchases a wrist watch for Rs. 450 and fixes its list price in such a way that after allowing a discount of 10%, he earns a profit of 20%. Then the list price (in rupees) of the wrist watch is.**

(a) 500

**(b) 600**

(c) 750

(d) 800

Sol.

If the marked price of the wrist watch be Rs. x, then

x * 90/100 = (450 * 120)/100 = 540

=> x = (540 * 100)/90 = Rs. 600

**Q2.The simple and compound interests on a sum of money for 2 years are Rs. 8400 and Rs. 8652 respectively. The rate of interest per annum is**

**(a) 6%**

(b) 7.5%

(c) 9%

(d) 4.5%

Sol.

Difference = Rs.(8652 – 8400)

= Rs. 252

so Rate = (2 * Difference)/S.I * 100

= (2 * (252/8400) * 100 = 6%

**Q3.A can complete a piece of work in 12 days. B is 60% more efficient than A. the number of days that B will take to complete the same work is**

(a) 6

**(b) 15/2**

(c) 8

(d) 17/2

Sol.

Ratio of their efficiency

= 100 : 160 = 5 : 8

so Ratio of time taken = 8 : 5

so Time taken by B

= 12 * 5/8 = 15/2days

**Q4.A boys was asked to find the value of 3/8 of a sum of money. Instead of multiplying the sum by 3/8. he divided it by 3/8 and then his answer exceeded by Rs. 55. Find the correct answer?**

(a)Rs. 9

**(b)Rs. 24**

(c)Rs. 64

(d)Rs. 1,320

Sol.

Let amount be Rs.x

so According to question,

8/3x – 3/8x = 55

x = Rs.24

**Q5.The least number which when divided by 4, 6, 8, 12 and 16 leaves a remainder of 2 in each case is:**

(a)46

(b)48

**(c)50**

(d)56

Sol.

L.C.M. of 4, 6, 8, 12, and 16 = 48

So Required number = 48+2 = 50

**Q6. When a number is first increased by 10% and then reduced by 10% the number:**

(a) Does not change

**(b) Decrease by 1%**

(c) Increase by 1%

(d) None of these

Sol.

Let the given number be x

Increased number = (110% of x)

=(110/100×x)=11x/10

Finally 11x/10, reduced number

=(90% of 11x/10)=(90/100×11x/10)=99x/100

Decrease =(x-99x/100)=x/100

Decrease % =(x/100×1/x×100 )%=1%

**Q7. Out of 100 students, 50 failed in English and 30 in Mathematics. If 12 students fail in both English and mathematics, then the number of students who passed in both the subjects is**

(a) 26

(b) 28

(c) 30

**(d) 32**

Sol.

Let A = set of students who fail in English and B = set of students who fail in mathematics

Then n(A) = 50, n(B) = 30 and n(A ∩B)=12

n(A∪B)=n(A)+n(B)-n(A∩B)=(50+30-12)=68

Number of students who fail in one or both the subjects = 68

Number of those who pass in both = (100 – 68) = 32

**Q8. 8% of the voters in an election did not cast their votes. In the election, there were only two candidates. The winner by obtaining 48% of the total votes defeated his contestant by 1100 votes. The total number of voters in the election were**

(a) 21000

(b) 22000

(c) 23500

**(d) 27500**

Sol.

Let the total number of voters be x

Votes cast = 92% of x =(92/100×x)=23x/25

Votes in favour of winning candidate =48/100×x=12x/25

Votes polled by defeated candidate

=(23x/25-12x/25)=11x/25

12x/25-11x/25=1100

⇒ 12x – 11x = 27500

⇒ x = 27500

**Q9. Because of scarcity of rainfall, the price of a land decreases by 12% and its production also decreases by 4%. What is the total effect on revenue?**

(a) Loss of 16%

(b) Gain of 15%

**(c) Loss of 15.52%**

(d) Gain of 15.48%

Sol.

Net effect =[-12-4+(-12)(-4)/100]% = (–16 + 0.48)% = –15.25%

**Q10. From 1980-1990, the population of a country was increased by 20%.**

**From 1990 – 2000, the population of the country was increased by 20%**

**From 2000-2010, the population of the country was increased by 20%**

**Then the overall increased population (in percentage) of the country from 1980-2010 was**

(a) 60%

(b) 62.8%

(c) 72.2%

**(d) 72.8%**

Sol.

Let the population in the year 1980 = x

Then, the population in the year 1990 = 120% of x

Then, the population in the year 2000

= 120% of 120% of x

And the population in year 2010

= 120% of 120% of 120% of x

=6/5×6/5×6/5×x=216/125 x

Hence, required percentage

=(216/125 – x)/x×100%

=91/125×100%=72.8%