**Question-1****Explanation**

**what will be the average mark of all the students if 80,90,70 are the average marks of the 3 groups of 40,50 and 60 students respectively?**

**Options :**

a)80

b)79

c)90

d)89

**Answer: ** ** b) 79**

From the given data

If 80 is the average marks of 40 students then 80*40=3200 is the total marks of 40 students.

If 90 is the average marks of 50 students then 90*50=4500 is the total marks of 50 students.

If 70 is the average marks of 60 students then 70*60=4200 is the total marks of 60 students

Now, the total number of students=(40+50+60)=150

Requred average={3200+4500+4200]/150=11900/150=79.33=79(approximately)

**Question-2****Explanation**

**Find the odd one-28,29,60,123,736,3685?**

**Options :**

a)29

b)123

c)736

d)3685

**Answer: ** ** b) 123**

The given series are in the following form,

28*1+1=29

29*2+2=60

60*3+3=183

183*4+4=736

736*5+5=3685

Instead of 183,123 is given in the series, so 123 is the odd one.

**Question-3****Explanation**

**Find the odd one-1853,925,461,229,113,55,11**

**Options :**

a) 229

b)113

c)55

d)11

**Answer: ** ** d)11**

Each number in the series is obtained by subtracting 3 and divide the result by 2.

1853-3/2=925

925-3/2=461

461-3/2=229

229-3/2=113

113-3/2=55

55-3/2=26

26-3/2=11.5

Clearly, 11 is wrong.

**Question-4****Explanation**

**Find the odd one-1900,1876,1855,1837,1821,1810,1801**

**Options :**

a)1855

b)1837

c)1821

d)1900

**Answer: ** ** c) 1821 **

The above series is formed by subtracting multiples of 3 starting from 24 in reverse order.

1900-24=1876

1876-21=1855

1855-18=1837

1837-15=1822

1822-12=1810…and so on..

The odd one is 1821 which has to be 1822.

**Question-5****Explanation**

**Veer invested an amount of Rs. 9000 for 2 years at compund interest rate 15% per annum. How much amount will Veer obtain as interest?**

**Options :**

a)Rs. 2902.50

b)Rs.2900.50

c)Rs.2899.50

d)Rs.2899

**Answer: ** ** a) Rs. 2902.50 **

When interest is compunded annually, we have to use the following formula:

Amount=P*[1+(R/100)]” where p=principal, R=rate of interest and n=time (years)

Here p=Rs.9000,R=15%,n=2 years.

Then amount=Rs.9000*[1+15/100)]^2=9000*

(23/20)^2=23805/2=Rs.11902.5

The amount obtained by the way of interest in compund interest=amount-principal=Rs.(11902.5-9000)=Rs.2902.50

Hence the required answer is Rs. 2902.50

**Question-6****Explanation**

**Shagi deposits Rs. 1500 each on 1st january and 1st july of a year at the rate of 8% compund interest calculated on half-yearly basis. How much amount he would have at the end of the year?**

**Options :**

a)Rs.2150.50

b)Rs.3140.40

c)Rs.3182.40

d)Rs.2152.50

**Answer: ** ** c) Rs. 3182.40**

When interest is compunded half-yearly: Amount=p*[1+(R/2)/100]^2

The total amount for the investment on 1st january is: amount1=Rs. 1500*[1+(8/2)/100^2*1=Rs.1500*[1+(4/100)]^2=Rs.1500*[26/25]^2

The total amount for investment on 1st july is: (Here n=1/2 year since it starts from 1st july to end of the same year)

Amount2=Rs. 1500*[1+(8/2)/100]^[2*(1/2)]
=Rs.1500*[1+4/100]
=Rs.1500*[26/25]
The total amount at the end of the year=amount1+amount2

=1500*[26/25]^2+1500*[26/25]
=1500*[26/25]*[(26/25)+1]
=1500*26/25*51/25

=3182.40

Hence Rs. 3182.40 is the required answer.

**Question-7****Explanation**

**What is the difference between the compund interests on Rs. 10,000 for 2 years at 5% per annum compunded yearly and half-yearly?**

**Options :**

a)Rs.6.00

b)Rs. 6.25

c)Rs. 6.50

d)Rs.6.75

**Answer: ** ** b) Rs. 6.25**

Here, p=Rs.10,000, n=2years, and R=5%

Amount invested for compund interest ( yearly)=Rs. 1000*[1+5/100]=Rs. 10000*21/20=Rs. 10,500

Amount invested for compund interest ( Half-yearly)=Rs. 10000*[1+(5/2)/100]^2=10000*(41/40)^2=Rs. 10,506.25

Difference =Rs. ( 10506.25-10500=Rs. 6.25

**Question-8****Explanation**

**What will come in the place of ? in the following question?
(88.128/?)/{(2.4*0.3)/(0.2*1.5)}=18**

**Options :**

a) 1.04

b)2.04

c)3.04

d)4.04

**Answer: ** ** b) 2.04**

Given that ( 88.128/?)/{(2.4*0.3)/(0.2*1.5)}=18

Let us take ? as X.

Then ,(88.128/x)/{(0.72)/(0.3)}=18

(88.128/x)/2.4=18

88.128/x=18*2.4

88.128/X=43.2

43.2*x=88.128

X=88.128/43.2

X=2.04

Hence the answer will be 2.04

**Question-9****Explanation**

**One-fifth of sqrt(X)*3=37.2-sqrt(12.96)**

**Options :**

a)2601

b)2916

c)3136

d)2704

**Answer: ** ** c) 3136**

Given that one –fifth os sqrt(X)*3=37.2-sqrt(12.96)

i.e.,1/5x sqrt(X)*3=37.2-3.6

3/5*sqrt(X)=33.6

Sqrt(X)=33.6*5/3

Sqrt(X)=11.2*5=56

X=562=3136

Hence the answer is 3136

**Question-10****Explanation**

**What will come in the place of X in the following question: 3.5 {2.5-[6.5-(1.4/X)]}*8.24/(0.2)2=813.08**

**Options :**

a)20

b)25

c)40

d)5.5

**Answer: ** ** a) 20**

3.5-2.5-(6.5(1.4/X))}*(8.24/0.04=813.08

3.5-(2.5-{(6.5X-1.4)/X)}*(824/4)=813.08

3.5-{(2.5X-6.5X+1.4)/X}*206=813.08

3.5-{(-4X+1.4)/X}*206=813.08

3.5-{(-824X-288.4)/X}=813.08

(3.5X+824X-288.4)/X=813.08

827.5X-288.4/X=813.08

827.5X-288.4=813.08X

827.5X-813.08X=288.4

14.42X=288.4

X=288.4/14.42

X=20