**1. Rs. 1500 were divided into two parts. One part was put at 6% and the other at 5% interest. If the whole annual interest from both investments was Rs. 85, then the investment at 6% was:**

A. Rs. 1200

**B. Rs. 1000**

C. Rs. 1300

D. Rs. 1150

**(Suppose, one part of sum = Rs. x**

**∴ Other part of the sum = Rs. (1500 – x)**

**According to question,**

**(x×6×1)/100+((1500-x)×5×1)/100=85**

**6x + 7500 – 5x = 8500 or, x = 1000)**

**2. 800 becomes Rs. 956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will Rs. 800 become in 3 years?**

A. Rs. 1020.80

B. Rs. 1025

**C. Rs. 1052**

D. Rs. 1050

**(SI = 956 – 800 = Rs. 156**

**Therefore, rate of interest**

**=(SI × 100)/(Principal × Time )**

**=(156 × 100)/(800 × 3)=6.5% per annum.**

**Thus, new rate = 10.5%**

**So,**

**S.I. =(Principal × Time × Rate )/100**

**=(800 × 3 × 105)/100 = Rs. 252**

**Hence, Amount = 800 + 252 = Rs. 1052.)**

**3. Prakash lends a part of Rs. 20,000 at 8% simple interest and remaining at 4/3% simple interest. His total income after a year was Rs. 800. Find the sum lent at 8%.**

**A. Rs. 8000**

B. Rs. 12000

C. Rs. 6000

D. Rs. 10000

**(Let the amount lent at 8% rate of interest be =Rs. x**

**∴ Amount lent at 4/3% rate of interest = Rs. (20,000 – x)**

**∴ SI =(Principal × Rate× Time )/100**

**∴(x × 8 × 1)/100+((20000-x) ×4/3 × 1)/100=800**

**⇒ 2x/25+(20000 – x)/75=800**

**⇒ (6x + 20000 – x)/75=800**

**⇒ 5x + 20000 = 75 × 800 = 60000**

**⇒ 5x = 60000 – 20,000 = 40000**

**⇒ x=40000/5 = Rs. 8000)**

**4. A man invested 1/3 of his capital at 7%, 1/4 at 8% and the remaining at 10% rate of simple interest. If his annual income from interests is Rs. 561, then the capital invested was:**

A. Rs. 6000

B. Rs. 5600

**C. Rs. 6600**

D. Rs. 7200

**(Let the total capital invested be Rs. x**

**∴ Total interest**

**=(1/3 x × 7 × 1)/100+(1/4 x × 8 × 1)/100+((1-1/3-1/4)x × 10x × 1)/100**

**=7x/300+8x/400+5x/120**

**=(28x + 24x + 50x )/1200=102x/1200**

**Now, according to the question,**

**561=102x/1200**

**∴ x=(561 × 1200)/102 = Rs. 6600)**

**5. Mohan can finish a job in 60 days whereas Ram can finish the same job in 20 days. If they work together, the job will be over in-**

A. 7.5 days

**B. 15 days**

C. 25 days

D. 55 days

**(1/t=1/60+1/20=1/15**

**Time =15 days)**

**6. A shopkeeper loses the S.P of 4 pencils on selling 36 pencils. His loss% –**

A. 12.5%

**B. 10%**

C. 15%

D. 25%

**(Let S.P of 1 pencil= 1 Rs , S.P of 36 pencils= 36 Rs**

**S.P<C.P (as loss) , C.P= 36+4= 40 Rs.**

**L%= (4/40) ×100= 10% )**

**7. If a, b, c are respectively the number of faces, edges and vertices of a pentagonal pyramid, then the value of (a-b+c/2)^2 – 2 is-**

A. 2

B. 1.75

**C. -1**

D. -1.5

**(Euler’s rule: F+V-E=2, a+c–b = 2**

**By putting it in question: (2/2)^2 – 2 = -1 ans)**

**8. An amount of Rs. 15,000 is distributed among A, B and C in the ratio 4:5:6. What is the share of B?**

**A. 5000**

B. 6000

C. 4000

D. 7000

**(let share of A=4x, B=5x, C=6x**

**Total amount = A+B+C ; 4x+5x+6x= 15000**

**15x= 15000, x= 1000**

**B’s share(5x) = 5000 Rs)**

**9. Find the total surface area of a hemisphere of radius 10 cm.**

**A. 942 cm square**

B. 547 cm square

C. 833 cm square

D. 978 cm square

**(T.S.A of hemisphere = 3πr^2 = 3 x 22/7 x 10 x 10**

**= 942 cm square)**

**10. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.**

A. 134 cm square

B. 126 cm square

C. 198 cm square

**D. 165 cm square**

**(C.S.A of the cone = πrl = 22/7 x 5.25 x 10**

**= 165 cm square)**