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# Maths Quiz for KVS and CTET Exams

Q1. At what rate of simple interest per annum will a sum become 7/4 of itself in 4 years?
(a) 18%
(b) 18 1/4%
(c) 18 3/4%
(d) 18 1/2%

Q2. What equal installment of annual payment will discharge a debt which is due as Rs. 848 at the end of 4 years at 4% per annum simple interest?
(a) Rs. 212
(b) Rs. 200
(c) Rs. 250
(d) Rs. 225

Q3. A certain sum amounts to Rs. 5,832 in 2 years at 8% per annum compound interest, the sum is:
(a) Rs. 5,000
(b) Rs. 5,200
(c) Rs. 5,280
(d) Rs. 5,400

Q4. What sum will give Rs. 244 as the difference between simple interest and compound interest at 10% in  1 1/2 years compounded half yearly?
(a) Rs. 40,000
(b) Rs. 36,000
(c) Rs. 32,000
(d) Rs. 28,000

Q5. In a mixture of 75 litres, the ratio of milk to water is 2 : 1. The amount of water to be further added to the mixture so as to make the ratio of the milk to water 1 : 2 will be
(a) 45 litres
(b) 60 litres
(c) 75 litres
(d) 40 litres

Q6. tan⁡θ/(1-cot⁡θ ) + cot⁡θ/(1-tan⁡θ ) is equal to
(a) 1 – tan θ – cot θ
(b) 1 + tan θ – cot θ
(c) 1 – tan θ + cot θ
(d) 1 + tan θ + cot θ

Q7. 2 cosec^2  23°cot^2 ⁡67°-sin^2 23°-sin^2 ⁡67°-cot^2 67  is equal to
(a) 1
(b) sec^2 ⁡23°
(c) tan^2⁡ 23°
(d) 0

Q8. The ratio of base of two triangles is x : y and that of their areas is a : b. Then the ratio of their corresponding altitudes will be:
(a) a/x ∶b/y
(b) ax : by
(c) ay : bx
(d) x/a ∶b/y

Q9. A playground is in the shape of a rectangle. A sum of Rs. 1,000 was spent to make the ground usable at the rate of 25 paise per sq. m. The breadth of the ground is 50 m. If the length of the ground is increased by 20 m. what will be the expenditure (in rupees) at the same rate per sq. m.?
(a) 1,250
(b) 1,000
(c) 1,500
(d) 2,250

Q10. A cuboidal water tank has 216 litres of water. Its depth is 1/3 of its length and breadth is 1/2 of 1/3 of the difference of length and breadth. The length of the tank is
(a) 72 dm
(b) 18 dm
(c) 6 dm
(d) 2 dm
Solutions

S2. Ans.(b)

Sol.
A=NI+(RI/100) x (N(N-1)/2)
A = Amount
N = Number of installment
R = Rate
848=4I+(4I/100)× 4(4-1)/2
848=4I+(4I/100)×6
848=4I+(24I/100)
848=(400I+24I)/100
848=424I/100
I = 200

S4. Ans.(c)
Sol. According to question,
R = 10% yearly
R=10/2%
= 5% monthly
Time = 1 1/2 year
= 3 half year
Difference =P(R/100)^2×((300 + R)/100)
244=P(5/100)^2×((300+5)/100)
244=P× (25/(100×100))×(305/100)
P = 32000

S9. Ans.(a)
Sol. Old expenditure = Rs. 1000
Increase in area = 50 × 20 m^2
Increase in expenditure = 50 × 20 × .25
= Rs. 250
⇒ New expenditure = 1000 + 250
= Rs. 1250

S10. Ans.(b)
Sol. Let l=9x,h=3x,b=x
l×b×h=216×1000
( 1 litre = 1000 cm^3)
9x × 3x × x = 216000
27x^3 = 216000
x^3 = 8000
x = 20
l = 180 cm = 18 dm

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