- A shopkeeper sells two articles-A & B. Cost price of article-B is 20% less than cost price of article-A and shopkeeper sells article-A and article-B at 40% profit and 20% profit respectively. If selling price of article-A is Rs.528 more than selling price of article-B, then find cost price of article B?
Rs. 1000Rs. 910Rs. 900Rs. 880Rs. 960Option E

Let cost price of article – A be Rs. 10x

So, cost price of article – B = 10x × 80/100 = Rs. 8x

And, Selling price of article – A = 10x × 140/100 = Rs. 14x

And selling price of article – B = 8x × 120/ 100 = Rs. 9.6 x

ATQ, 14x – 9.6x = 528

4.4x = 528

x = Rs.120

Hence, cost price of article – B = 8x = Rs. 960 - Area of circle is 144π cm^2 and radius of circle is equal to diagonal of a square. Find perimeter of square.
20√5 cm18√3 cm24√2 cm12√2 cm20√3 cmOption C

Area of circle = πr²

ATQ, πr² = 144π

⇒ r = 12cm

Let side of a square be ‘a’ cm.

So, a² + a² = (12)²

2a² = 144

a² = 72

a = 6√2 cm

So, required perimeter = 4a = 24√2 cm - Ayush invested Rs.5000 in a scheme-A on S.I. for two years and he further invested the amount received from scheme-A on C.I. at the rate of 10% compounding annually for two years. If he received Rs.1218 as C.I., then find rate of interest of scheme-A?
18%12%8%10%14%Option C

Let rate of interest offered by scheme – A be R% p.a.

Amount invested by Ayush at C.I = 5000 × R × 2/100 + 5000 = (100R + 5000) Rs.

Equivalent rate of interest of 10% C.I. for two years = 10 + 10 + (10 × 10)/100 = 21%

ATQ, [(100 R+5000) × 21]/100 = 1218

⇒ 21R + 1050 = 1218

⇒ R = 8% - A vessel of capacity of 70 liters is completely filled with milk. ‘x’ liters are taken out from the vessel and replaced by water and this process is repeated one more time. If final quantity of milk in the vessel is 44.8 liters, then find value of ‘x’?
1612101418Option D

44.8 = 70 (1 − x /70)^2

16/25 = (1 – x/ 70)^2

So, x = 14, 126 lit.

As x cannot be greater than 70 litres

So, x =14 lit.

So, 14 liters of mixture can be taken out as capacity of vessel is only 70 liters. - In 2018, a school has 1200 students and ratio of boys to girls is 11 : 9. If 92% of the total students got passed in 2018 and 95% of the total boys got passed in 2018 then, find the percentage of girls who got passed in 2018?
74(1/3)%89(1/5)%80(1/2)%88(1/3)%81(1/2)%Option D

Total number of students who got passed in 2018 = 1200 × 92/100 = 1104

Total number of boys who got passed in 2018 = 1200 × 11/20 × 95/100 = 627

Required % = (1104 − 627)/1200 × 9/20 × 100

= 477/540 × 100 = 265/3 = 88(1/3)% - 𝟐𝟑. 𝟖𝟑% 𝐨𝐟 𝟔𝟐𝟓. 𝟎𝟐 − 𝟏𝟎𝟎. 𝟎𝟏 =? % 𝐨𝐟 𝟑𝟓𝟗𝟗. 𝟗𝟗 + 𝟗𝟖. 𝟏𝟑 ÷ 𝟔. 𝟗𝟗𝟗
32146Option C

𝟐𝟑. 𝟖𝟑% 𝐨𝐟 𝟔𝟐𝟓. 𝟎𝟐 − 𝟏𝟎𝟎. 𝟎𝟏 =? % 𝐨𝐟 𝟑𝟓𝟗𝟗. 𝟗𝟗 + 𝟗𝟖. 𝟏𝟑 ÷ 𝟔. 𝟗𝟗𝟗

24/100 × 625 – 100 = ?/100 × 3600 + 98/7

⇒ 50 = ? × 36 + 14

⇒ ? = (50 – 14)/36

⇒ ? = 1 - 𝟔𝟓𝟗. 𝟗𝟕 × (? )^𝟐 = (𝟔𝟒. 𝟗𝟐) ^𝟐 + 𝟐𝟒. 𝟗𝟗𝟕% 𝐨𝐟 𝟔𝟖𝟔𝟎. 𝟎𝟎𝟏
31245Option A

𝟔𝟓𝟗. 𝟗𝟕 × (? )^𝟐 = (𝟔𝟒. 𝟗𝟐) ^𝟐 + 𝟐𝟒. 𝟗𝟗𝟕% 𝐨𝐟 𝟔𝟖𝟔𝟎. 𝟎𝟎𝟏

660 ×(? )^2 = (65)² + 25/100 × 6860

=>660 × (? )^2= 4225 + 1715

=>(? )^2= 5940/660

=>? = √9

=> ? = 3 - 𝟓𝟔𝟕𝟕. 𝟏𝟑𝟐𝟏 + 𝟒𝟗𝟏𝟑. 𝟗𝟏𝟑𝟑 − 𝟑𝟕𝟗𝟖. 𝟗𝟐 = ? +𝟐𝟎. 𝟎𝟎𝟓% 𝐨𝐟 𝟑𝟗𝟔𝟎. 𝟏𝟑𝟐1
50004000300020006000Option E

𝟓𝟔𝟕𝟕. 𝟏𝟑𝟐𝟏 + 𝟒𝟗𝟏𝟑. 𝟗𝟏𝟑𝟑 − 𝟑𝟕𝟗𝟖. 𝟗𝟐 = ? +𝟐𝟎. 𝟎𝟎𝟓% 𝐨𝐟 𝟑𝟗𝟔𝟎. 𝟏𝟑𝟐1

5677 + 4914 – 3799 = ? + 20/100 × 3960

=>6792 = ? + 792

=> ? = 6000 - ? = 𝟓𝟔. 𝟗𝟏𝟓𝟔 𝐨𝐟 𝟐𝟖. 𝟎𝟓𝟔 ÷ 𝟕𝟔. 𝟎𝟕𝟓𝟒 × 𝟓. 𝟗𝟕𝟒
112126139150130Option B

? = 𝟓𝟔. 𝟗𝟏𝟓𝟔 𝐨𝐟 𝟐𝟖. 𝟎𝟓𝟔 ÷ 𝟕𝟔. 𝟎𝟕𝟓𝟒 × 𝟓. 𝟗𝟕𝟒

=>? = 57 × 28 × 1/76 × 6

=>? = 126 - 𝟏𝟏. 𝟗𝟗𝟒% 𝐨𝐟 𝟓𝟎𝟎. 𝟎𝟑 + 𝟏𝟔.𝟎𝟏×?/ 𝟐𝟎.𝟎𝟒 = 𝟏𝟓𝟎. 𝟎𝟏𝟐𝟑 + (𝟐𝟓. 𝟗𝟓𝟑𝟏 × 𝟑𝟓. 𝟏𝟐𝟏)
11221330125010001200Option C

𝟏𝟏. 𝟗𝟗𝟒% 𝐨𝐟 𝟓𝟎𝟎. 𝟎𝟑 + 𝟏𝟔.𝟎𝟏×?/ 𝟐𝟎.𝟎𝟒 = 𝟏𝟓𝟎. 𝟎𝟏𝟐𝟑 + (𝟐𝟓. 𝟗𝟓𝟑𝟏 × 𝟑𝟓. 𝟏𝟐𝟏)

12/100 × 500 + (16 × ?)/20 = 150 + (26 × 35)

=> 60 + (4 × ?)/ 5 = 150 + 910

=>4 × ?/5 = 1060 – 60

=>? = (1000 × 5)/ 4

=>? = 1250

**Directions(6-10):** What approximate value will come in place of question mark ‘?’ in the following questions.

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