Q1) A man sells sugar to a tradesman at a profit of 30%, but tradesman, becoming bankrupt pays only 64 paise per rupee. How much percent does the man gain or lose by his sale?

A.16.8 % gain

B. 16.8 % loss 

C. 13.6 gain 

D. 13.6 loss
 

Answer : Option B

Solution : To get a fix, we shall assume that the man sells the sugar of cost price Rs 100 at a profit of 30%, that is he sells the sugar for Rs 130. But the tradesman goes bankrupt and he pays only 64 paise in a rupee owed by the tradesman.

So, instead of paying Rs. 130, the tradesman pays only Rs (130 × 64)/100 = Rs 83.2.

Loss suffered by the dealer selling sugar to the tradesman = Rs 100 — Rs 83.2 = Rs 16.8 loss

 loss = ( 16.8/100) × 100 = 16.8%.

Q2) If the average marks of three batches of 55,60 and 45 students respectively is 50,55 and 60, then the average marks of all the students is

1. 53.2
2. 54.68
3. 55.55
4. 56.72

Answer : Option B

Solution : Average marks of 3 batches of 55,60 and 45 students are 50,55 and 60 respectively.

Total students =50+60+45=160

Marks total =(55×50)+(60×55)+(45×60)

                   =2750+6000

                   =8750

Average marks = 8750/16 =54.6875.

Q3) Ratio of lemon to water to be mixed to prepare lemon juice is 3:5. For 12 lemons the number of cups of water to be used is ?

A. 10

B. 15

C. 20

D. 25

Answer : Option C

Solution :Ratio of Lemon to water=3:5

Number of lemons =12

3 parts of lemon =12

1 part of lemon = 12/3 = 4

5 parts of water ≡4×5=20

number of cups of water to be used=20

Q4) Subhadra and Janaki together finish the job in 24 days. Kartik who is half as efficient as Subhadra joins them and they finish the job in 20 days. How many days will it take for Subhadra and Kartik to finish the same job? 

A. 40 days

B. 20 days

C. 60 days

D. 100 days
 

Answer : Option A

Solution : S + J = 24 days

K + S + J = 20 days.

Chocolates = LCM (24, 20) = 120C

Efficiency - S + J = 120C/24d = 5 c/d

Efficiency – K +S + J = 120C/20d = 6 c/d

So, K efficiency = 1c/d

K is half efficient as S,

Then Efficiency of S = 2 c/d

So, efficiency of S + K = 3 c/d.

Days by S + k = 120c/(3c/d) = 40 days

Q5) The ratio of areas of square and circle is given n : 1 where n is a natural number. If the ratio of side of square and radius of circle is k :1, where k is a natural number, then n will be multiple of

A. 7

B. 22

C. 154

D. None

Answer : Option C

A.12.50%

B.10%

C.8.50%

D.14%

Answer : Option A

Solution : 

Let Principle = P.

Given, Amount = 2P

n = 8 years.

I = ?

SI = PNR/100.

SI = A-P

SI = 2P-P =P

So, P = P8R/100,

R = 100/8 = 12.5%

Q7) A stone is dropped from a height of 5 km. The distance it falls through varies directly with the square of the time taken to fall through that distance. If it falls 64 m in 4 seconds, find the distance the stone covers in the 5th second?

A.36 m

B.58 m

C.72 m

D.100 m

Answer : Option A

Solution : 

A stone is dropped from a height of 5 km.

The distance it falls through varies directly with the square of the time

Let us suppose distance as d and time as t

d

d

where, k is constant of proportionality

If a stone travels 64 m in 4 second.

d=64 and t=4

Substitute into the formula and solve for k

64=16k

k=4 m/s

We get an relation between d and t

d-4t²

We need to find distance covered by stone in the 5th second.

Distance covered in 5th second = d(5)-d(4)

At t=5, d=4*5^2 = 100 m

At t=4, d=64m

Distance covered in 5th second = 100-64 = 36 m

Q8) What is the difference between the LCM and HCF of the numbers 20, 30, and 40?

A.100

B.110

C.120

D.130

Answer : Option B

Solution : LCM(20, 30, 40) – HCF(20, 30, 40) = 120-10 = 110.

Q9) log 3600 is equal to:

A.2 log 6 + 1

B.6 log 2 + 1

C.2 log 6 + 2

D.6 log 2 +2

Answer : Option C

Solution : Log3600 = log(36*100) = log36 + log100

= log 6^2 + log10^2

= 2log6 + 2 log10

= 2log6 + 2.1

= 2log6 + 2

Q10) Find the probability that the sum of the score is even in a throw of two dice.

A.1/4

B.1/3

C.1/2

D.2/3

Answer : Option C

Solution : Probability that the sum is even when throwed 2 dice = Favourable occurrences/total occurrences

Totally 36 sums that are possible, out of which 18 are even and 18 are odd.

So, probability of even sums = 18/36 = 1/2

Q11) Riya earns Rs. 30,000 p.m. She spends 10% of her income on taxes. Of the remaining, she spends one-third on rent, half the remaining on petrol, one third of the remaining amount on electricity and the rest goes in the savings. What percentage of Riya's income is she able to save?

A.5%

B.10%

C.15%

D.20%

Answer : Option D

Solution : Riya’s earning = 30000.

Tax expense = 10%30000 = 3000.

Remaining amount = 27000.

Rent expense = 1/3(27000) = 9000.

Remaining amount = 18000.

Petrol expense = 1/2(18000) = 9000

Remaining = 9000.

Electricity = 1/3(9000) = 3000.

Remaining is the Rest = Saving = 6000.

X%30000 =6000

X = 20%

Q12) How many factors does 400 have?

A.8

B.10

C.12

D.15

Answer : Option D

Solution : 

400 = 4*100 = 2²* 10²

       = 2² * (2*5)² = 2⁴ * 5²

So, 400 = 2⁴ * 5²

No. of factors = (P+1)*(P+1) = (4+1)*(2+1) = 15.

Q13) If I have kept six different books on a shelf, in how many different ways can I arrange them?

A.6

B.24

C.120

D.720

Answer : Option D

Solution : 6 books, 6 places on shelf = 6! Ways = 720 ways

Q14) A bag contains coins of 1 rupee, 2 rupees and 5 rupees in the ratio 4:8:5. The total amount in the bag is Rs. 90. The number of 5 rupee coins is:

A.8

B.16

C.15

D.10

Answer : Option D

Solution : 

No. of 1-rupee coins = 4x

No. of 2-rupee coins = 8x

No. of 5-rupee coins = 5x.

Given total amount in the bag = Rs. 90

4x coins each of 1 rupee, the total value of 1-rupee coins = 4x*1

8x coins each of 2 rupee, the total value of 2 rupee coins = 8x*2

5x coins each of 5 rupee, the total value of 5 rupee coins = 5x*5

So, 4x + 16x + 25x = 90

45x = 90.

x = 2.

Total number of 5-rupee coins = 5x = 5*2 = 10 coins

Q15) A box contains 5 red, 4 white and 3 green balls. In how many ways can 3 balls be drawn from the box, without replacement, so that at least 2 of them are green?

A.18

B.28

C.27

D.9

E.30

Answer : Option B

Solution : 

Total 3 balls to be selected and at least 2 of them are green.

There are 5 red, 4 white and 3 green from which the required balls to be picked.

Different ways = 2 green and 1 red OR 2 green and 1 white OR 3 green

3C2*5C1 or 3C2*4C1 or 3C3 = (3*5) + (3*4) + 1 = 28