## SOLVED PROBLEM

(TPSC - Small Savings held on 4th April,2010)

**Q1** **A's** salary is 20% below of **B's** salary. By how much percent is B's salary is above A's ?

**Sol :** Let **B's** salary is 100.

Therefore, **A's** salary is 80.

Therefore, **B's** salary is above **A's** salary is ^{20}⁄_{80} X 100%

= 25%

Ans : = 25%

**Q2.** P's money is Q's money as 4 : 5 and Q's money is R's money as 2 : 3. If P has 800, R has

**Sol :** P:Q = 4 : 5

⇒ ^{P}⁄_{Q} = ^{4}⁄_{5} (Given P = 800)

⇒ ^{800}⁄_{Q} = ^{4}⁄_{5}

⇒ Q = ^{5 x 800}⁄_{4} = 1000

Again, Q : P = 2 : 3

⇒ ^{Q}⁄_{P} = ^{2}⁄_{3} (Here Q = 1000)

⇒ ^{1000}⁄_{P} = ^{2}⁄_{3}

⇒ P = ^{3 x 1000}⁄_{2} = 1500

Ans: P = 1500

**Q3.** At what rate percent per year simple interest will a sum be doubled in 8 years?

**Sol:** Let, Principal is 100 , therefore after 8 years a sum 200 i.e. simple interest 100.

In 8 years 100 rupees simple interest 100

^{100}⁄

_{8}

= 12

^{4}⁄

_{8}

= 12

^{1}⁄

_{2}

Ans: 12

^{1}⁄

_{2}

**Q4**. A tap can fill in 25 minutes and another tap can empty in 50 minutes. If both are opened simultaneously, when will be the tank full?

**Sol : **Fill in 25 minutes i.e. 100% but same time empty will be 50% as the time taken double to empty.

So, remain in tank 50%. To fill more 50% more 25 minutes time required.

i.e. total time 25 + 25 minutes = 50 minutes.

Ans : 50 minutes.

**Q5.** The cost of paving a rectangular courtyard of 30 m long and 20 m wide with tiles of 12 cm long and 10 cm wide at Rs. 40 per thousand tiles is

**Sol :** 30 m = 30 x 100 cm = 3000 cm and 20 m = 20 x 100 cm = 2000 cm

Area of courtyard = length x bredth

= 3000 x 2000 sq. cm

= 6000000 sq. cm

Area of tiles = length x bredth

= 12 x 10 sq. cm

= 120 sq. cm

Total tiles required =
Area of courtyard ÷ Area of tiles

= ^{6000000}⁄_{120} Nos.

= 50000 nos.

Cost per thousand is Rs. 40

Here 50 thousand tiles require

Therefore, Total cost 50 x 40 = 2000

Ans : 2000

**Q6.** In a cricket match, average number of runs scored by five players comes to 25. One of them scored 13 rans. What is the average score of remaining players?

**Sol :** Total run of five players = 25 x 5 = 125

One scored 13, so remaining four’s total run = 125 – 13 = 112

Therefore, average run = ^{112}⁄_{4} = 28

Ans : 28

**Q7.** Which of the following is the greatest?

^{4}⁄

_{5},

^{7}⁄

_{13},

^{11}⁄

_{15}

**Sol**: Here LCM is 195, So, multiply by other factor to get denominator 195 and by the the same factor multiply the numerator.

i.e ^{4}⁄_{5} is greatest.

**More..**