Maths Questions for CTET,KVS Exam : 21 november 2018

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. If 56M4 is completely divisible by 11, then what is the value of M?
(a) 0

(b) 1
(c) 3 
(d) 5

Q2. A and B can together do a piece of work in 10 days. If A works with twice of his efficiency and B works with an efficiency 1/3rd less than his efficiency,then the work gets completed in 6 days. In how many days can A and B do the work alone respectively?
(a) 40/3, 40

(b) 20/3, 20
(c) 30, 20/3
(d) 50/3, 25

Q3. In the given figure, ABCD is a rectangle. F is a point on AB and CE is drawn perpendicular to DF. If CE = 60cm and DF = 40cm, then what is the area (in cm2) of the rectangle ABCD?

(a) 1200 

(b) 1800
(c) 2400
(d) 2800

Q4. What will be the net discount (in %) of two successive discounts of 15% and 35%?
(a) 44.75 

(b) 51.25
(c) 55.25 
(d) 56.25

Q5. A company, at the time of inflation reduced the staff in the ratio 5 : 3 and average salary per employee is increased in the ratio 7 : 8. By doing so, the company saved Rs 55000. What was the initial expenditure 
(in Rs) of company?

(a) 155000 

(b) 160000
(c) 175000 
(d) 215000

Q6. a, b and c are 3 values, such that a + b = 5, b + c = 7.5 and c + a = 8.5. What will be the average of these values?
(a) 1.5 

(b) 3
(c) 3.5 
(d) 4.5

Q7. Due to increase of 33.33% in the price of apples, a customer can purchase 4 apples less for Rs 16. What is the original price (in paise) of an apple?
(a) 100 

(b) 125
(c) 150 
(d) 400/3

Q8. Due to increase of k% in the side, the area of a square increases by 69%. What is the value of k?
(a) 30 

(b) 33
(c) 34.5 
(d) 35

Q9. A starts from a point at a speed of 30 metre/second. After 3 seconds, B starts chasing A from the same point with a speed of 50 metre/second. What will be the total distance (in metres) travelled by A and B before A is caught by B?
(a) 360 

(b) 450
(c) 600 
(d) 720

Q10. The difference between compound interest and simple interest on a sum for 2 year at 20% per annum is Rs 200. If the interest is compounded half yearly, then what is the difference (in Rs) between compound and simple interest for 1st year?
(a) 50 

(b) 75
(c) 100 
(d) 150

Solution

S1. Ans.(d); For divisibility by 11,
Sum of digits at even places minus sum of digits at odd places should be either divisible by 11 or 0
∴ 6 + 4 – 5 – M = 0 or a multiple of 11
If value of M is 5, then it will be divisible by 11.

S2. Ans.(a); Let A can do in a days.
And B can do in b days.
ATQ,
1/a+1/b=1/10 …(i)
2/a+2/3b=1/6…(ii)
Multiplying equation (i) by 2 and subtracting equation (ii) from (i)
∴ b = 40
& a=40/3

S3. Ans.(c);

Join FC
Area of triangle DFC=1/2×40×60 =1200 cm^2
Now Area of triangle DFC is also
=1/2×DC×BC
So,
1200 × 2 = DC × BC
DC × BC = Area of rectangle = 2400 cm².

S4. Ans.(a); Let SP be 100
After 1st discount
SP will be = 85
After 35% (2nd) discount
S.P =85×65/100=55.25
∴ Net discount = 100 – 55.25
= 44.75

S5. Ans.(c); Let the initial number of employee be 5x
And final number of employee be 3x
Therefore,
Total initial salary = 7 × 5x = 35x
And Total final salary = 8 × 3x = 24x
∴ ATQ,
35x – 24x = 55000
11x = 55000
x = 5000
∴ initial expenditure of company = 35 × 5000 = Rs. 175000

S6. Ans.(c); Adding all
2(a + b + c) = 5 + 7.5 + 8.5 = 21
(a+b+c)/3=21/6=3.5

S7. Ans.(a); 33.33%=(+1)/3
∴ Total Initial apples = 4 × 4 = 16
∴ Original price=16/16=1 Rs.


S8. Ans.(a); Let side of square be 10 unit.
∴ ATQ,
((100+K)/10)^2–100=169 –100
((100)^2+200K+K^2)/100–100=69
K² + 200K – 6900 = 0
⇒ K = 30 or –230
∴ K = 30

S9.Ans. (b); Distance covered by A in 3 seconds = 30 × 3 = 90 metre.
This 90 m distance is covered by B in
=90/(50-30)=4.5 second
∴ A travelled = 90 + 4.5 × 30
= 90 + 135 = 225 m
& B travelled = 4.5 × 50 = 225 m
∴ Required total distance = 225 + 225 = 450 m