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# Maths Questions for CTET,KVS Exam : 3rd october 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. A car goes one kilometre at 30 km per hour and then goes another kilometre at 40 km per hour. The average speed (in km/hr) of the car for 2 km is
(a) 35
(b) 342/7
(c) 333/7
(d) 335/7
Q2. Find the HCF of x³ + 64 and x² – 4x + 16.
(a) x² – 4x + 16
(b) 1
(c) x – 4
(d) x + 4
Q3. The expression 1 – p² – 2q + q² is factorized as
(a) (p + q – 1) (q – q – 1)
(b) (p + q + 1) (p + q – 1
(c) (p + q + 1) (p – q + 1
(d) (p – q – 1) (p – q + 1)
Q4. Factorise x⁶ – 5x⁴ + 6x².
(a) (x² – 3) (x² – 2)
(b) x²(x² – 3)(x² + 2)
(c) x²(x² – 3)(x² – 2)
(d) x²(x² + 3)(x² – 2)
Q5. If xy = 6 and x²y + xy² + x + y = 63, then the value of x² + y² is
(a) 23
(b) 55
(c) 61
(d) 69
Q6. The present age of Ramesh’s father is thrice the age of Ramesh. After 5 years, the difference of their ages will be 30 years. Find the present age of Ramesh.
(a) 15
(b) 25
(c) 20
(d) 45
Q7. A sum of money amounts to Rs. 4,818 after 3 years and Rs. 7, 227 after 6 years on compound interest. The sum is
(a) Rs. 3,212
(b) Rs. 2,409
(c) Rs. 2,490
(d) Rs. 3,122
Q8. A scale on a map is 1:10000. What actual distance does a length of 8 cm on the map represent?
(a) 0.8 km
(b) 2 km
(c) 1.6 km
(d) 4 km
Q9. If the cost price of 10 candle is equal to the selling price of 8 candles, the gain/loss percent is
(a) 25% gain
(b) 20% loss
(c) 20% gain
(d) 25% loss
Q10. Aaron deposited Rs. 1000 at a rate of 4% (p.a.) for 73 days. Find the amount he got back after the completion of the time period.
(a) Rs. 1008
(b) Rs. 1080
(c) Rs. 8000
(d) Rs. 1800
Solutions S2. Ans.(a)
Sol. Let f (x) = x³ + 64 and g (x)
= x² – 4x + 16
f (x) = x³ + 64 = x³ + 4³
= (x + 4) (x² – 4x + 16)
g (x) = x² – 4x + 16
Since x² – 4x + 16 is common in f (x) and g (x).
HCF = x² – 4x + 16
S3. Ans.(a)
Sol. After rearranging the terms of the given expression, 1 – p² – 2q + q² = q² – 2q + 1² – p²
= (q – 1)² – p²
As we know that a² – b²
= (a + b) (a – b), (q – 1)² – p²
= (q – 1 + p) (q – 1 – p)
= (p + q – 1) (q – p – 1)
S4. Ans.(c)
Sol. x⁶ – 5x⁴ + 6x²
= x⁶ – 3x⁴ – 2x⁴ + 6x²
= x⁴(x² – 3) – 2x²(x² – 3)
= (x² – 3)(x⁴ – 2x²)
= x²(x² – 3) (x² – 2)
S5. Ans.(d)
Sol. Putting the value of xy = 6 in x²y + xy² + x + y = 63, we get
6x + 6y + x + y = 63
7x + 7y = 63
x + y = 9
Now, x² + y² = (x + y)² – 2xy
= 9² – 2 × 6
= 81 – 12 = 69
S6. Ans.(a)
Sol. Let the present age of Ramesh be x years,
∴ Present age of his father = 3x
After 5 years,
Age of Ramesh = (x + 5) years
Age of his father = (3x + 5) years
According to the question,
(3x + 5) – (x + 5) = 30
⇒ 3x + 5 – x – 5 = 30
⇒ 2x = 30
⇒ x = 15
∴ The present age of Ramesh is 15 years. S8. Ans.(a)
Sol.
(Distance on map)/(Actual distance)=1/10000
⇒ Actual distance = Distance on map × 10000
Actual distance = 8 × 10000 cm
= 0.8 km
S9. Ans.(a)
Sol. The cost price of 10 candles is equal to the selling price of 8 candles.
Let the cost price of 10 candles = Selling price of 8 candles = x
∴ Cost price of 1 candle = x/10
And the selling price of 1 candle = x/8
Profit on one candle = x/8-x/10
=(5x-4x)/40
=x/40
Gain %=(x/40×100)/(x/10)%=25%
S10. Ans.(a)
Sol. Time = 73 days
=73/365 year=1/5 year
SI=PRT/100
=(1000×4×1)/(100×5)
= Rs. 8
A = P + SI = 1000 + 8
= Rs. 1,008