**Directions(1-5):** What value will come in place of question mark ‘?’ in the following questions.

- 36% of 245 − 40% of 210 = 10 − ?
4.43.26.55.82.9Option D

36% of 245 − 40% of 210 = 10 − ?

=>? = 10 – 88.2 + 84 = 5.8 - 3( 1/7 ) + 2( 3/5) + 𝟕( 1/5) – 5(3/7)− 18/35 = 35/?
37254Option D

3( 1/7 ) + 2( 3/5) + 𝟕( 1/5) – 5(3/7)− 18/35 = 35/?

35/? = 263/35 – 18/35

=> ? = 35×35/245 = 5 - √1225 ÷ (343)^1/3 × 45% of 760 = ?
17271820189319891710Option E

√1225 ÷ (343)^1/3 × 45% of 760 = ?

? = 35 ÷ 7 × 342 = 1710 - 175% of 460 + 110% of 170 + 2^? = 1000
72153Option E

175% of 460 + 110% of 170 + 2^? = 1000

=>2^? = 1000 − 805 − 187 = 8

=> ? = 3 - 18^7.9 × 3^0.1 × 6^0.1 ÷ (3^4 × 6^4) = 18^?
42315Option A

18^7.9 × 3^0.1 × 6^0.1 ÷ (3^4 × 6^4) = 18^?

=>18^? = 18^7.9 × 18^0.1 ÷ 18^4

=> ? = 4 - Cost price of article A is double then that of article B and shopkeeper mark up both the article 20% more than the cost price. If at the time of sale shopkeeper gave Rs. 9 discount and earn 17% profit on total. Find the cost price of article A?
263368289345200Option E

Let C.P. of A = 2x

C.P. of B = x

Total cost price = 3x

Mark up price = 3x × 12/10 = 3.6x

ATQ, 3.6x – 9 = 3x × 1.17

⇒ 3.6x – 3.51x = 9

⇒ 0.09x = 9

⇒ x = 100

C.P. of article A = 200 - Ravi invested Rs 18000 in scheme ‘A’ which offers 15% p.a. at simple interest and Rs 15000 in scheme ‘B’ which offers 18% p.a at compound interest. Find the difference between the interest earn from these two schemes after two years?
400440450452486Option E

Interest earn from scheme ‘A’ = 18000 × 15 × 2/100 = 5400

Interest earn from Scheme ‘B’ = 15000 [(1 + 18/100)^2 − 1] = 15000 [ 3924/10000] = 5886

Required difference = 5886 – 5400 = 486 - A tank can be filled with water by two pipes A and B together in 36 minutes. If the pipe B was closed after 30 minutes, the tank is filled in 40 minutes. The pipe B can alone fill the tank in?
5070608090Option E

In 30 min the part of the tank will be filled by both tap = 30/36 = 5/6

Required tap = 1 – 5/6 = 1/6

1/ 6 part of the tank will be filled by tank A in 10 min.

∴ tap A will take 60 min.

∴ tap B will take time to fill the tank = 1/36 – 1/60 = 1/90

Required time = 90 min. - 40 men, working 8 hours a day can do a piece of work in 15 days. Find the number of days in which second group of 60 men working 4 hrs a day can do twice the work. Assume that 3 men of the first group do as much work in 2 hour as 4 men of the second group do in 3 hrs.
70608050100Option C

Let efficiency of men of first group is M1 and second group is M2

3M1 × 2 = 4M2 × 3

M1 = 2M2

work = 40M1 × 8 × 15

According to question,

40M1 × 8 × 15 × 2 = 60M2 × 4 × d

d = 80 days - An alloy of copper and aluminum has 40% copper. Another alloy of Copper and Zinc has Copper and Zinc in the ratio 2: 7. These two alloys are mixed in ratio 5:3. Quantity of aluminum is what percent more/less than the quantity of copper in final alloy.
14.4%13.8%12.5%11.6%10.9%Option C

Let quantity → 5x and 3x

Quantity of copper in final alloy = 2/5 × 5x + 2/9 × 3x = 8x/3

Quantity of Aluminium = 3/5 × 5x = 3x

Required percentage = [3x – 8x/3]/ 8x/3 × 100 = 12.5%