**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. tanθ/(1-cotθ ) + cotθ/(1-tanθ ) 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