Maths Questions for CTET,KVS Exam : 15 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. PQRS is a cyclic quadrilateral and PQ is the diameter of the circle. If RPQ = 38°, then what is the value (in degrees) of PSR?

(a) 52
(b) 77
(c) 12
(d) 142

Q2. Smaller diagonal of a rhombus is equal to length of its sides. If length of each side is 6 cm, then what is the area
(in cm2) of an equilateral triangle whose side is equal to the bigger diagonal of the rhombus?

(a) 18√3
(b) 27√3
(c) 32√3
(d) 36√3

Q3. In the given figure, PR and ST are perpendicular to tangent QR, PQ passes through center O of the circle whose diameter is 10cm. If PR = 9cm, then what is the length (in cm) of ST?

(a) 1
(b) 1.25
(c) 1.5
(d) 2

Q4. Production of company A is 120% of the production of company B and 80% of the production of company C. What is the ratio between the productions of companies A, B and C respectively? 
(a) 6 : 5 : 9
(b) 6 : 5 : 4
(c) 12 : 10 : 15
(d) 10 : 12 : 15

Q5. A, B and C together start a business. The ratio of investment of A and B is 7 : 8 and that of B and C is 4 : 9. B gets a share of Rs. 7104 in annual profit. What is C’s share in the profit? 
(a) Rs. 16984
(b) Rs.16894
(c) Rs. 15984
(d) Rs. 14894

Q6. If Suresh sells an article at a price of Rs. 9300, he gets a profit of Rs. 3100. At what price should he sell the article so that he gets a profit of 25%? 
(a) Rs. 7250
(b) Rs. 7350
(c) Rs. 7650
(d) Rs. 7750

Directions (7-10): The table given below shows the percentage of literate people in 6 cities. The table also shows the ratio of males to females among literate people. 

% of literate people of any city
= (Literate people of the city/Total population of the city) × 100 

Q7. If the total population of City 4 is 600000, then how many literate people are there in city 4?
(a) 480000
(b) 378000
(c) 468000
(d) 348000


Q8. Total population of City 6 is 200000 and the total population of City 2 is 220000. What is the respective ratio of literate males of City 2 and literate females of City 6?
(a) 348 : 595
(b) 255 : 199
(c) 595 : 348
(d) 199 : 255

Q9. If there are 259210 literate females in City 5, then what is the total population of City 5?
(a) 644000
(b) 354200
(c) 690000
(d) 483000

Q10. The population of the 6 cities are 250000, 200000, 220000, 300000, 150000 and 400000 respectively. Which is the correct order of the number of literate people in these cities?
(a) City 6 > City 1 > City 4 > City 2 > City 3 > City 5
(b) City 4 > City 6 > City 1 > City 2 > City 3 > City 5
(c) City 6 > City 4 > City 1 > City 3 > City 2 > City 5
(d) City 6 > City 1 > City 4 > City 3 > City 2 > City 5

Solutions

S1. Ans.(c)
Sol. 

Given than,
∠RPQ = 38°
∠PRQ = 90° due to diameter angle in circle
∠RQP = 180 – (90 + 38) = 52°
∠PSR = 180° – 52 = 128° (Quadrilateral opposite angle property)

S2. Ans.(b)
Sol. 

OC² = 36 – 9
OC = 3√3
AC = 6√3
Area of ∆ACB
=√3/4×6√3×6√3 = 27 √3

S3. Ans.(a)
Sol. 

OP = OS = 5 cm
PR = 9 cm, RN = 5 cm
Then PN = 9 – 5 = 4 cm
ON = 3 cm due to triplet.
∆OPN ≈ ∆SLO,
∠SLO = ∠ONP = 90° &
∠PON ≈ ∠OSL ⇒ similar ∆ triangle
PN = OL = 4 cm ⇒ LM = ST = 5 – 4 = 1 cm

S4. Ans.(c)
Sol. Let the production of company B = 100 units
∴ Production of company A = 120 units
Production of company C
= (120×100)/80=150 units
∴ Required ratio
=120∶100∶150=12∶10∶15

S5. Ans.(c)
Sol. A : B = 7 : 8
B : C = 4 : 9 = 8 : 18
∴ A : B : C = 7 : 8 : 18
Sum of ratios = 7 + 8 + 18 = 33
Now, C’s share = 18/8×7104
= Rs. 15984

S6. Ans.(d)
Sol. C.P. of article
=Rs.(9300-3100)
=Rs.6200
To gain 25%
S.P. = (6200×125)/100
= Rs. 7750

S7. Ans.(b)
Sol. Given then,
Literate people in city 4
=600000/100×63 =378000

S8. Ans.(c)
Sol. Req.ratio=(220000/100×85/11×7)/(200000/100×58/5×3) =1190/696=595/348

S9. Ans.(a)
Sol. Total population of city 5
=[259210/7×16]/92×100 = 644000

S10. Ans.(d)
Sol. City 1 = 80% of 250000 = 200000
City 2 = 85% of 200000 = 170000
City 3 = 78% of 220000 = 171600
City 4 = 63% of 300000 = 189000
City 5 = 92% of 150000 = 138000
City 6 = 58% of 400000 = 232000
So, city 6 > city 1 > city 4 > city 3 > city 2 > city 5