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# Maths Quiz for KVS/NVS and CTET Exams Q1. If two circles are such that the centre of each lies on the circumference of the other, then the ratio of the common chord of two circles to the radius of any of the circle is:
(a) √3 ∶2
(b) √3 ∶1
(c) √5 ∶1
(d) None of these

Q2. If ∛a+∛b=∛c, then the simplest value of (a+b-c)^3+27abc is
(a) -3
(b) 0
(c) -1
(d) 3

Q3. A container of 64 liter was full of milk. 16 liter milk was drawn off and replaced by water and this process is repeated two times more. Find quantity of water in the mixture now.
(a) 37 liter
(b) 27 liter
(c) 21 liter
(d) 36 liter

Q4. In what time will Rs. 10000 amount to Rs. 13310 at 20% per annum compounded half yearly?
(a) 1 1/2 years
(b) 2 years
(c) 2 1/2 years
(d) 3 years

Q5. A sum of Rs. 2525 is lent out in two parts in such a way that the interest on one part at 7% for 4 years is equal to that on another part at 6% for 7 years. Find the difference between the two parts of the sum.
(a) Rs. 505
(b) Rs. 1515
(c) Rs. 1010
(d) None of these

Q6. A tradesman gives 4% discount on the market price and gives 1 article free for buying every 15 articles and thus gains 35%. The marked price is increased above the cost price by:
(a) 20%
(b) 39%
(c) 40%
(d) 50%

Q7. A profit of 12% is made mobile phone is sold at Rs. P and there is 4% loss when the phone is sold at Rs. Q. Then Q : P is
(a) 4 : 5
(b) 3 : 1
(c) 1 : 1
(d) 6 : 7

Q8. From a container of wine, a thief has stolen 15 litres of wine and replaced it with same quantity of water. He again repeated the same process. Thus in three attempts the ratio of wine and water became 343:169. The initial amount of wine in the container was:
(a) 150 litres.
(b) 120 litres.
(c) 135 litres.
(d) 125 litres.

Q9. A student walks from his house at 5/2 km/h and reaches his school late by 6 min. Next day, he increase his speed by 1 km/h and reaches 6 min before school time. How far is the school from his house ?
(a) 5/4 km
(b) 7/4 km
(c) 9/4 km
(d) 11/4 km

Q10. A car run 1 km in 4 min 50 sec and B in 5 minutes. How many metres’ start can A give B in a km race so that the race may end in a dead heat?
(a) 33.33m.
(b) 25m.
(c) 20m.
(d) 40m.

Solutions

S2. Ans.(b)
Sol.
∛a+∛b=∛c
=∛a+∛b-∛c=0  …(i)
If a + b + c = 0
Then a^3+b^3+c^3=3abc
From equation (i)
a+b-c=-3∛abc

Cubing both side
(a+b-c)^3=-27 abc
(a+b-c)^3+27abc=0

S4. Ans.(a)
Sol. 20% yearly = 10% half year
=11/10
13310/10000=(11/10)^n
(11/10)^3=(11/10)^n
n = 3 years
=1 1/2  year

S6. Ans.(d)
Sol.
Let the C.P. of each article be Rs. x
Then, C.P. of 16 articles = Rs. 16x
S.P. of 15 articles = 135% of Rs. 16x
=Rs.(135/100)×16x=Rs.108x/5
S.P. of 1 article = Rs. (108x/5)×(1/15) = Rs. 36x/25
If S.P. is Rs. 96, M.P. = Rs. 100
If S.P. is Rs. =36x/25, M.P. = Rs. (100/96)×(36x/25)
= Rs. 3x/2
Increase in M.P. over C.P.
=((3x/2)-x)/x)×100
= 50%

S7. Ans.(d)
Sol.
Let the cost of mobile phone = 100
So, P = 112 Rs.
Q = 96 Rs.
Required ratio Q : P = 96 : 112 = 6 : 7

S9. Ans.(b)
Sol.
According to the formula,
Required distance = ab(t1+t2 )/(b – a)
Where a = 5/2, b = 7/2, t1 = 6, t2 = 6
∴ Required distance = ( (5/2) × (7/2) x (12/60) ) / (7/2) – ( 5/2)
(35/4) x (1/5)
=7/4  km

S10. Ans.(a)
Sol.
Time taken by A to run 1 km = 290 seconds.
Time taken by B to run 1 km = 300 seconds.
A can give B a start of (300 – 290) sec, i.e. 10 sec.
In 300 sec, B runs 1000 m.
In 10 sec, B runs (1000/300) × 10 m = 100/3 m
∴ A can give B a start of 100/3 m.