**Dear Students!!! There is most general as well as a scoring section in all the competitive entrance examinations in the teaching field i.e “Mathematics”.Because in this section only one thing is work i.e your accuracy and that could be nourished with the daily practice. So, for this, we are providing you the daily quiz for all teaching exams i.e CTET Exam 2018, DSSSB ,KVS,STET Exam.**

**Q1. How many numbers are there between 1 to 200 which are divisible by 3 but not by 7?**

(a) 38

(b) 45

(c) 57

(d) 66

**Q2. 10 women can do a piece of work in 6 days, 6 men can do same work in 5 days and 8 children can do it in 10 days. What is the ratio of the efﬁciency of a woman, a man and a child respectively?**

(a) 4 : 6 : 3

(b) 4 : 5 : 3

(c) 2 : 4 : 3

(d) 4 : 8 : 3

**Q3. The ratio of the volume of two cylinders is 7 : 3 and the ratio of their heights is 7 : 9. If the area of the base of the second cylinder is 154 cm2, then what will be the radius (in cm) of the ﬁrst cylinder?**

(a) 6√2

(b) 6√3

(c) 7√2

(d) 7√3

**Q4. Kanchan bought a clock with 25% discount on marked price. She sold it with 75% gain on the price she bought. What was her proﬁt percentage on the marked price?**

(a) 31.25

(b) 50

(c) 56.25

(d) 60

**Q5. A, B and C received an amount of Rs 8400 and distributed among themselves in the ratio of 6 : 8 : 7 respectively. If they save in the ratio of 3 : 2 : 4 respectively and B saves Rs 400, then what is the ratio of the expenditures of A, B and C respectively?**

(a) 6 : 8 : 7

(b) 8 : 6 : 7

(c) 9 : 14 : 10

(d) 12 : 7 : 9

**Q6. The average age of 24 students is 12 years. It was observed that while calculating the average age, the age of a student was taken as 14 years instead of 8 years. What will be the correct average age (in years)?**

(a) 11.25

(b) 11.5

(c) 11.75

(d) 12.25

**Q7. 70% of the cost price of an article is equal to the 40% of its selling price. What is the proﬁt or loss percentage?**

(a) 63% loss

(b) 70% loss

(c) 75% proﬁt

(d) 80% proﬁt

**Q8. a% of b + b% of a = ______**

(a) 2a% of b

(b) 2a% of 2b

(c) 2a% of 2a

(d) 2b% of 2b

**Q9. If I walk at 7/6 of my usual speed, then I reach my ofﬁce 15 minutes early. What is the usual time taken (in minutes) by me to reach the ofﬁce?**

(a) 60

(b) 75

(c) 90

(d) 105

**Q10. A person lent Rs 10000 to B for 3 years and Rs 6000 to C for 4 years on simple interest at same rate of interest and received Rs 5400 in all from both of them as interest. What is the rate of interest (in %)?**

(a) 10

(b) 12.5

(c) 15

(d) 20

Solutions

S1. Ans.(c)

Sol. Total no. which are divisible by 3 = 66

Total no. which are divisible by 3 but not by 7 means which are not divisible by 21= 9

∴ Total number which are divisible by 3 but not by 7 = 66 – 9 = 57

S2. Ans.(d)

Sol. Ratio of total work of women, men and children

= 10×6 : 6×5 : 8×10

= 60 : 30 : 80

∴ Ratio of efficiency

=1/60 ∶1/30 ∶1/80

= 4 : 8 : 3

S3. Ans.(d)

Sol. Let the radius of first and second cylinder be r₁ and r₂ and height be h₁ and h₂ respectively.

∴ ATQ,

(πr_1^2 h_1)/(πr_2^2 h_2 )=7/3

(r_1^2×7)/(r_2^2×9)=7/3⇒(r_1^2)/(r_2^2 )=3⇒r_1=√3 r_2

r_2=r_1/√3…(i)

πr_2^2=154…(ii)

putting value of equation (i) in (ii)

22/7×(r_1^2)/3=154

r₁ = 7√3 cm

S4. Ans.(a)

Sol. Let marked price be Rs. 100.

Therefore, cost price = Rs.75

Selling price=75×175/100=Rs.131.25

∴ Required profit percentage

=(131.25 –100)/100×100

=31.25 %

S5. Ans.(c)

Sol. Amount received by A

=8400×6/21=2400

Amount received by B

=8400×8/21=3200

Amount received by A

=8400×7/21=2800

Saving of A=400/2×3=600

Saving of B = 400

Saving of C=400/2×4=800

Therefore ratio of expenditure

= (2400 – 600) : (3200 – 400) : (2800 – 800)

= 1800 : 2800 : 2000

= 9 : 14 : 10

S6. Ans.(c)

Sol. Total age of 24 student = 24 × 12 = 288

Correct total = 288 – 6 = 282

∴ Correct average=282/24

=11.75

S7. Ans.(c)

Sol. 70% of CP = 40% of SP

SP/CP=7/4

∴ Let CP be 4x and SP be 7x

∴Profit percent=(7x–4x)/4x×100

=3x/4x×100=75%

S8. Ans.(a)

Sol. a/100×b+b/100×a=2ab/100

= 2a% of b

S9. Ans.(d)

Sol. Let the usual speed be 6x km/hr

Therefore, speed at which he reaches 15 min. early is 7x km/hr

∴ ATQ,

d/6x–d/7x=15/60

d/x [1/42]=1/4

d/x=21/2

∴ distance (d)=21/2 x km

∴ usual time=(21/2 x)/6x

=7/4 hours=105 minutes.

S10. Ans.(a)

Sol. ATQ,

(10000×3×R)/100+(6000×4×R)/100=5400

[where R is rate of interest]

30R + 24R = 540

∴R=540/54=10%