41. Two numbers are 20% and 25% respectively lesser than a third number. By how much percent is the second number to be enhanced to make it equal to the first number?

दो संख्याएँ एक तीसरी संख्या से क्रमशः 20% और 25% कम हैं। पहली संख्या के बराबर करने के लिए दूसरी संख्या को कितने प्रतिशत तक बढ़ाया जा सकता है?

Option “B” is correct.

CALCULATION:

Let the third number be = 100

⇒ 1st number is 20% less = 80

⇒ 2nd number is 25% less = 75

Second number to be enhanced by = (80 – 75)/75 × 100%

∴ Required percentage = 6.67%

42. A student who scored 18% marks in (10 + 2) examination and failed by 20 marks but another student who got 34% marks got 12 marks more than the pass marks in the examination. Then find the percentage of marks that are required to pass in the examination.

एक छात्र ने (10 + 2) परीक्षा में 18% अंक प्राप्त किए और 20 अंकों से फेल हो गया, लेकिन 34% अंक प्राप्त करने वाले दूसरे छात्र को परीक्षा में उत्तीर्ण अंकों से 12 अंक अधिक प्राप्त हुए। तो परीक्षा में उत्तीर्ण होने के लिए आवश्यक अंकों का प्रतिशत ज्ञात कीजिये।

Option “B” is correct.

Given:

Percent of marks obtained by first student = 18% 

The first student failed by 20 marks

Percent of marks obtained by second student = 34%

The second student gain more marks = 12 marks 

Formulae used:

Total marks = (Sum of marks/Difference in % marks) × 100

Calculations:

Let the total marks be x 

Marks obtained by the first student = x × (18/100) = 9x/50

Passing marks = (9x/50) + 20      —-(1)

Marks obtained by second student = x × (34/100) = 17x/50

Passing marks = (17x/50) – 12      —-(2)

Now, equating equation (1) and (2)

(9x/50) + 20 = (17x/50) – 12 

⇒ 8x/50 = 32 

⇒ x = 200

Passing marks = (9x/50) + 20

⇒ 9 × (200/50) + 20

⇒ 56 

Percentage of passing which is required to pass = (56/200) × 100 = 28%

∴ Percentage of passing which is required to pass is 28%

 

Alternate Method:

Total marks = (Sum of marks/Difference in % marks) × 100

⇒ (32/16) × 100 = 200

Passing marks = (9x/50) + 20

⇒ 9 × (200/50) + 20

⇒ 56 

Percentage of passing which is required to pass = (56/200) × 100 = 28%

∴ Percentage of passing which is required to pass is 28%

43. Price of the laptop is increased by 25%. Now the second time what percent of price increases such that the total increase becomes 35%?

Option “C” is correct.

Given:

1) Price is increased by 25% = x

2) Total increased by 35%

Formula:

(1) Total increase = x + y + xy/100

Calculation:

⇒ 35 = 25 + y + (25 × y/100)

⇒ 35 = 25 + y + y/4

⇒ 140 = 100 + 4y + y

⇒ 140 – 100 = 5y

⇒ 40 = 5y

⇒ y = 8

∴ Second time increment is 8%.

44. If the length of a rectangle increases by 50% and the breadth decreases by 25%, then what will be the percent increase in its area?

यदि एक आयत की लंबाई में 50% की वृद्धि और चौड़ाई में 25% की कमी होती है, तो इसके वर्तमान क्षेत्रफल में कितनी वृद्धि होगी?


Option “C” is correct.

GIVEN:

Length of a rectangle increases by 50%

Breadth decreases by 25%

CONCEPT:

Taking both length and breath as 100 will help in calculation.

FORMULA USED:

Area = length × breath

CALCULATION:

Let length and breadth be 100 each.

Area = 10000

After change in length and breadth

→ Length = 150, Breadth = 75

Area = 150 × 75 = 11,250

→ % increase in area = ((11,250 – 10,000)/10,000) × 100

→ % increase in area = 12.5%

45. A worker’s salary was increased by 30%, so that his salary became Rs. 910. How much did he earn before the increase?

Option “C” is correct.

Assume salary = x

Given :

After increased by 30%, it became 910 Rs.

∴ x × 130/100 = 910

⇒ x = 700 Rs.

46. If a side of a rectangle is increased by 20% and its breadth is decreased by 10%. Then find the percentage change in the area.

यदि एक आयत की एक भुजा में 20% की वृद्धि और इसकी चौड़ाई में 10% की कमी की जाती है। तो इसके क्षेत्रफल में प्रतिशत परिवर्तन ज्ञात कीजिये।

Given:

Percentage change in length = 20% increase

Percentage change in breadth = 10% decrease

Concept used:

Area of rectangle = Length × Breadth

Calculation:

Let the length of the rectangle = x unit

And the breadth of the rectangle = y unit

⇒ Area of rectangle (A) = xy unit2

Inceased length = x + 20% of x = 1.2x unit

Decreased breadt = y – 10% of y = 0.9y unit

⇒ Area of changed rectangle (A’) = 1.2x × 0.9y = 1.08xy unit2

⇒ Change in area = A’ – A = 1.08xy – xy = 0.08xy unit2

Percentage change in area = (Change in area/A) × 100

⇒ Percentage change in aea = (0.08xy/xy) × 100

∴ Percentage change in area = 8%

Shortcut Trick

By using successive formula for percentage change = (x + y + xy/100)%

Here x = 20 and y = –10

⇒ Percentage change in area = (20 – 10 – (20 × 10)/100)%

∴ Percentage change in area is 8%

47.Income of A is 25% more than the income of B. If income of A is increased in the ratio 5 ∶ 6 while the income of B increased in the ratio 4 ∶ 5. Then what percent income of A is more than the income of B?

A की आय B की आय से 25% अधिक है। यदि A की आय में 5 : 6 के अनुपात में वृद्धि होती है, जबकि B की आय में 4 ∶ 5 के अनुपात में वृद्धि होती है, तो A की आय B की आय से कितने प्रतिशत अधिक है?


Option “B” is correct.

Let the income of B be 100x.

Income of A = 100x × (125/100) = 125x

After increment income of A = 125x × 6/5 = 150x

After increment income of B = 100x × 5/4 = 125x

Income of A is more than the income of B by = (150x – 125x/125) × 100 = 20%

48. The price of a certain food items was raised by 12% in market. The consumption of the same food items was decreased from 250 kgs to 200 kgs. By how much percent will expenditure on food item fall in the market?

बाज़ार में किसी विशेष खाद्य पदार्थ की कीमत 12% से बढ़ गई। इसी खाद्य पदार्थ की खपत 250 किग्रा से घटकर 200 किग्रा हो गई। बाज़ार में खाद्य पदार्थों पर व्यय कितने प्रतिशत से गिर जाएगा?

Option “A” is correct.

Let initial price of food be Rs. 100

As known,

Expenditure = price × consumption

For price 100 consumption is 250 kgs

Expenditure = 100 × 250 = 25000.

Price of a food items is increased by 12%.

New price of food item = 100 + 100 × 12/100 = 112

New expenditure = 112 × 200 = 22400.

Change in expenditure 

= [(25000 – 22400)/25000] × 100

= 10.4

∴ By 10.4% expenditure of food item is falled.[/bg_collapse]

49. The passing marks of an exam is 60% marks. A student gets 320 marks and yet fails by 16 marks. What are the maximum marks?

एक परीक्षा मे उत्तीर्णता अंक 60% है। एक छात्र 320 अंक प्राप्त करता है फिर भी 16 अंक से अनुत्तीर्ण हो जाता है। अधिकतम अंक क्या हैं?


Option “D” is correct.

Let the maximum marks be a.

According to the question,

⇒ (60a/100) = 320 + 16

⇒ a = 336/0.6 = 560

∴ Maximum marks are 560.

50. 72% of a number is 90. What is the number?

यदि, किसी संख्या का 72 प्रतिशत 90 है, तो वह संख्या ज्ञात कीजिये।

Option “B” is correct.

Let a number be ‘x’,

72% of a number is 90,

⇒ x × 72/100 = 90

⇒ x = 125

∴ Number is 125.

 

Pages: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

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